The tangent of angle between the lines
Question: The tangent of angle between the lines whose intercepts on the axes are $a,-b$ and $b,-a$, respectively, is A. $\frac{a^{2}-b^{2}}{a b}$ B. $\frac{b^{2}-a^{2}}{2}$ c. $\frac{b^{2}-a^{2}}{2 a b}$ D. None of these Solution: C. $\frac{b^{2}-a^{2}}{2 a b}$ Explanation: Let the first equation of line having intercepts on the axes $a,-b$ is $\frac{x}{a}+\frac{y}{-b}=1$ $\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=1$ $\Rightarrow \mathrm{bx}-\mathrm{ay}=\mathrm{ab}...
Read More →A wire of length 20 m is to be cut into two pieces.
Question: A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum? Solution: Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length $x$ and $y$, respectively. Then, $x+y=20$ .... (1) Perimeter of square, 4 (Side) $=x$ $\Rightarrow$ Side $=\frac{y}{3}$ Area ...
Read More →Find the value
Question: Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $9 x^{2}+16 y^{2}=144$ Solution: Given: $9 x^{2}+16 y^{2}=144$ Divide by 144 to both the sides, we get $\frac{9}{144} x^{2}+\frac{16}{144} y^{2}=1$ $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ (i) Since, $169$ So, above equation is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (ii) Comparing eq. ...
Read More →A wire of length 28 m is to be cut into two pieces.
Question: A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum? Solution: Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of length $x$ m and $y$ m, respectively. Then, $x+y=28$ ......(1) Perimeter of square, 4 (side) $=x$ $\Rightarrow$ Side $=\frac{x}{4}$ $\Rightarrow$ Are...
Read More →The equation of the line passing through the point (1, 2)
Question: The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 isA. y x + 1 = 0B. y x 1 = 0C. y x + 2 = 0D. y x 2 = 0 Solution: B. y x 1 = 0 Explanation: Given that line passing through the point (1, 2) And perpendicular to the line x + y + 1 = 0 Let the equation of line L is x y + k = 0 (i) Since, L is passing through the point (1, 2) 1 2 + k = 0 ⇒k = 1 Putting the value of k in equation (i), we get x y + 1 = 0 Or y x 1 = 0 Hence, the correct opt...
Read More →The equation of the straight line passing
Question: The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x isA. x y = 5B. x + y = 5C. x + y = 1D. x y = 1 Solution: B. x + y = 5 Explanation: Given that straight line passing through the point (3, 2) And perpendicular to the line y = x Let the equation of line L is y y1= m(x x1) Since, L is passing through the point (3, 2) y 2 = m(x 3) (i) Now, given eq. is y = x Since, the above equation is iny = mx + bform So, the slope of this equation is ...
Read More →Slope of a line which cuts off intercepts
Question: Slope of a line which cuts off intercepts of equal lengths on the axes isA. 1B. 0C. 2D. 3 Solution: A. 1 Explanation: We know that the equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$ Where $\mathrm{a}$ and $\mathrm{b}$ are the intercepts on the axis. Given that $\mathrm{a}=\mathrm{b}$ $\Rightarrow \frac{x}{a}+\frac{y}{a}=1$ $\Rightarrow \frac{x+y}{a}=1$ $\Rightarrow x+y=a$ $\Rightarrow y=-x+a$ $\Rightarrow y=(-1) x+a$ Since, the above equation is in $\mathrm{y}=\mathr...
Read More →Find the value
Question: Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $x^{2}+4 y^{2}=100$ Solution: Given: $x^{2}+4 y^{2}=100$ Divide by 100 to both the sides, we get $\frac{1}{100} x^{2}+\frac{4}{100} y^{2}=1$ $\frac{x^{2}}{100}+\frac{y^{2}}{25}=1$ (i) Since, $10025$ So, above equation is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (ii) Comparing eq. (i)...
Read More →A beam is supported at the two ends and is uniformly loaded.
Question: A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by (i) $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$ (ii) $M=\frac{W_{x}}{3} x-\frac{W}{3} \frac{x^{3}}{L^{2}}$ Find the point at which M is maximum in each case. Solution: (i) Given : $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$ $\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-2 \times \frac{W x}{2}$ $\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-W x$ For maximum or minimum values of $M$, we...
Read More →A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
Question: A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by (i) $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$ (ii) $M=\frac{W_{x}}{3} x-\frac{W}{3} \frac{x^{3}}{L^{2}}$ Find the point at which M is maximum in each case. Solution: (i) Given : $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$ $\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-2 \times \frac{W x}{2}$ $\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-W x$ For maximum or minimum values of $M$, we...
Read More →A line cutting off intercept – 3 from the y-axis
Question: A line cutting off intercept 3 from the y-axis and the tangent at angle to the x-axis is3/5, its equation is A. 5y 3x + 15 = 0 B. 3y 5x + 15 = 0 C. 5y 3x 15 = 0 D. None of these Solution: A. 5y 3x + 15 = 0 Explanation: Given that $\tan \theta=\frac{3}{5}$ We know that, Slope of a line, $m=\tan \theta$ $\Rightarrow$ Slope of line, $\mathrm{m}=\frac{3}{5}$ Since, the lines cut off intercepts $-3$ on $y$-axis then the line is passing through the point $(0,-3)$. the point $(0,-3)$. So, the...
Read More →Find the value
Question: Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $16 x^{2}+25 y^{2}=400$ Solution: Given: $16 x^{2}+25 y^{2}=400$ Divide by 400 to both the sides, we get $\frac{16}{400} x^{2}+\frac{25}{400} y^{2}=1$ $\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$ (i) Since, $254$ So, above equation is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (ii) Comparing e...
Read More →If p is the length of perpendicular from the origin
Question: If $p$ is the length of perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$ and $a_{2}, p_{2}, b_{2}$ are in A.P, then show that $a_{4}+b_{4}=0$. Solution: Given equation is $\frac{x}{a}+\frac{y}{b}=1$ Since, $p$ is the length of perpendicular drawn from the origin to the given line $\therefore \mathrm{p}=\left|\frac{\frac{0}{\mathrm{a}}+\frac{0}{\mathrm{~b}}-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}}}\right|$ Squaring both the sides, we have $\mathr...
Read More →Of all the closed cylindrical cans (right circular),
Question: Of all the closed cylindrical cans (right circular), which enclose a given volume of $100 \mathrm{~cm}^{3}$, which has the minimum surface area? Solution: Let $r$ and $h$ be the radius and height of the cylinder, respectively. Then, Volume $(V)$ of the cylinder $=\pi r^{2} h$ $\Rightarrow 100=\pi r^{2} h$ $\Rightarrow h=\frac{100}{\pi r^{2}}$ Surface area $(S)$ of the cylinder $=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r \times \frac{100}{\pi r^{2}}$ $\Rightarrow S=2 \pi r^{2}+\frac{200...
Read More →Find the equations of the lines through
Question: Find the equations of the lines through the point of intersection of the lines x y + 1 = 0 and 2x 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5. Solution: Given two lines are $x-y+1=0 \ldots \ldots$ (i) And $2 x-3 y+5=0 \ldots$ (ii) Now, point of intersection of these lines can be find out as: Multiplying equation (i) by 2 , we get Multiplying equation (i) by 2, we get $2 x-2 y+2=0 \ldots \ldots$ (iii) On subtracting equation (iii) from (ii), we get $2 x-2 y+2-2 x+3 y-5=0$...
Read More →Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.
Question: Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum. Solution: Let the two numbers be $x$ and $y .$ Then, $x+y=15$ ......(1) Now, $z=x^{2} y^{3}$ $\Rightarrow z=x^{2}(15-x)^{3}$ $[$ From eq. $(1)]$ $\Rightarrow \frac{d z}{d x}=2 x(15-x)^{3}-3 x^{2}(15-x)^{2}$ For maximum or minimum values of $z$, we must have $\frac{d z}{d x}=0$ $\Rightarrow 2 x(15-x)^{3}-3 x^{2}(15-x)^{2}=0$ $\Rightarrow 2 x(15-x)=3 x^{2}$ $\Rightarrow 30 x-2 x^{2}=3 x...
Read More →Find the value
Question: Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$ Solution: Given: $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$ (i) Since, $4936$ So, above equation is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (ii) Comparing eq. (i) and (ii), we get $a^{2}=49$ and $b^{2}=36$ $\Rightarrow a=\sqrt{49}$ and $b=\sqrt{36}$ ...
Read More →Find the value
Question: Find the (i) lengths of major axes (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ Solution: Given: $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ (i) Since, 25 9 So, above equation is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ ...........(ii) Comparing eq. (i) and (ii), we get $a^{2}=25$ and $b^{2}=9$ $\Rightarrow a=\sqrt{2} 5$ and $b=\sqrt...
Read More →Find the equation of the line which passes through
Question: Find the equation of the line which passes through the point ( 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point. Solution: Let $A B$ be a line passing through a point $(-4,3)$ and meets $x$-axis at $A(a, 0)$ and $y$-axis at $B(0, b)$ Using the section formula for internal division, we have $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}...
Read More →How should we choose two numbers,
Question: How should we choose two numbers, each greater than or equal to -2whose sum______________ so that the sum of the first and the cube of the second is minimum? Solution: Let the two numbers be $x$ and $y$. Then, $x, y-2$ and $x+y=\frac{1}{2}$ .....(1) Now, $z=x+y^{3}$ $\Rightarrow z=x+\left(\frac{1}{2}-x\right)^{3}$ [From eq. (1)] $\Rightarrow \frac{d z}{d x}=1+3\left(\frac{1}{2}-x\right)^{2}$ For maximum or minimum values of $z$, we must have $\frac{d z}{d x}=0$ $\Rightarrow 1+3\left(\f...
Read More →How should we choose two numbers,
Question: How should we choose two numbers, each greater than or equal to-2, whose sum______________ so that the sum of the first and the cube of the second is minimum? Solution: Let the two numbers be $x$ and $y$. Then, $x, y-2$ and $x+y=\frac{1}{2}$ .....(1) Now, $z=x+y^{3}$ $\Rightarrow z=x+\left(\frac{1}{2}-x\right)^{3}$ [From eq. (1)] $\Rightarrow \frac{d z}{d x}=1+3\left(\frac{1}{2}-x\right)^{2}$ For maximum or minimum values of $z$, we must have $\frac{d z}{d x}=0$ $\Rightarrow 1+3\left(\...
Read More →A straight line moves so that the sum
Question: A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point. Solution: We know that intercepts form of a straight line is $\frac{x}{a}+\frac{y}{b}=1$ Where $a$ and $b$ are the intercepts on the axes Given that $\frac{1}{a}+\frac{1}{b_{k}}=\frac{1}{k}$ (let) On cross multiplication we get $\Rightarrow \frac{k}{a}+\frac{k}{b}=1$ This shows that the line is passing through the fixed point ( $k, k$ )...
Read More →In what direction should a line be drawn through the point (1, 2)
Question: In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance 6/3 from the given point. Solution: Let the given line $x+y=4$ and the required line ' $T$ ' intersect at $B(a, b)$ Slope of line ' $T$ ' is $\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{\mathrm{b}-2}{\mathrm{a}-1}$ And we also know that, $m=\tan \theta$ $\therefore \tan \theta=\frac{\mathrm{b}-2}{\mathrm{a}...
Read More →Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
Question: Divide 64 into two parts such that the sum of the cubes of two parts is minimum. Solution: Suppose64isdividedintotwopartsxand64-x.Then, $z=x^{3}+(64-x)^{3}$ $\Rightarrow \frac{d z}{d x}=3 x^{2}+3(64-x)^{2}$ For maximum or minimum values of $z$, we must have $\frac{\mathrm{dz}}{\mathrm{dx}}=0$ $\Rightarrow 3 \mathrm{x}^{2}+3(64-\mathrm{x})^{2}=0$ $\Rightarrow 3 x^{2}=3(64-x)^{2}$ $\Rightarrow x^{2}=x^{2}+4096-128 x$ $\Rightarrow x=\frac{4096}{128}$ $\Rightarrow x=32$ Now, $\frac{d^{2} z...
Read More →Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
Question: Divide 64 into two parts such that the sum of the cubes of two parts is minimum. Solution: Suppose64isdividedintotwopartsxand64-x.Then, $z=x^{3}+(64-x)^{3}$ $\Rightarrow \frac{d z}{d x}=3 x^{2}+3(64-x)^{2}$ For maximum or minimum values of $z$, we must have $\frac{\mathrm{dz}}{\mathrm{dx}}=0$ $\Rightarrow 3 \mathrm{x}^{2}+3(64-\mathrm{x})^{2}=0$ $\Rightarrow 3 x^{2}=3(64-x)^{2}$ $\Rightarrow x^{2}=x^{2}+4096-128 x$ $\Rightarrow x=\frac{4096}{128}$ $\Rightarrow x=32$ Now, $\frac{d^{2} z...
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