A window in the form of a rectangle is surmounted by a semi-circular opening.
Question: A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening. Solution: Let the dimensions of the rectangular part be $x$ and $y$. Radius of semi-circle $=\frac{x}{2}$ Total perimeter $=10$ $\Rightarrow(x+2 y)+\pi\left(\frac{x}{2}\right)=10$ $\Rightarrow 2 y=\left[10-x-\pi\left(\frac{x}{2}\right)\right]$ $\Rightarrow y=\frac{1}...
Read More →A tank with rectangular base and rectangular sides,
Question: A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is $2 \mathrm{~m}$ and volume is $8 \mathrm{~m}^{3}$. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank? Solution: Let $l, b$ and $h$ be the length, breadth and height of the tank, respectively. Height, $h=2 \mathrm{~m}$ Volume of the tank $=8 \mathrm{~m}^{3}$ Volume of the tank $=l \times b \t...
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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. $16 x^{2}+y^{2}=16$ Solution: Given: $16 x^{2}+y^{2}=16$ Divide by 16 to both the sides, we get $\frac{16}{16} x^{2}+\frac{1}{16} y^{2}=1$ $\frac{x^{2}}{1}+\frac{y^{2}}{16}=1$ (i) Since, $116$ So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (ii) Comparing eq. (i) and (...
Read More →A tank with rectangular base and rectangular sides,
Question: A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is $2 \mathrm{~m}$ and volume is $8 \mathrm{~m}^{3}$. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank? Solution: Let $l, b$ and $h$ be the length, breadth and height of the tank, respectively. Height, $h=2 \mathrm{~m}$ Volume of the tank $=8 \mathrm{~m}^{3}$ Volume of the tank $=l \times b \t...
Read More →If the latus rectum of an ellipse is equal
Question: If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity. Solution: Equation of an ellipse $=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Whereas Length of latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$ Length of minor axis $=2 b$ $\mathrm{SO}$ $\frac{2 b}{2}=\frac{2 b^{2}}{a}$ (Given) $A b=2 b^{2}$ $2 b^{2}-a b=0$ $b(2 b-a)=0$ So, $b=0$ or $a=2 b$ $b^{2}=a^{2}\left(1-e^{2}\right)$ $\mathrm{b}^{2}=(2 \mathrm{~b})^{2}\left(1-\mathrm{e}^{2}\right)$...
Read More →Find the equation of a circle concentric
Question: Find the equation of a circle concentric with the circle x2+ y2 6x + 12y + 15 = 0 and has double of its area. Solution: Given equation of the circle is x2 6x + y2+ 12y + 15 = 0 The above equation can be written as x2 2 (3) x + 32+ y2+ 2 (6) y + 62+ 15 9 + 36 = 0 (x 3)2+ (y-(-6))2 30 = 0 (x 3)2+ (y-(-6))2=(30)2 Since, the equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2 Centre = (3,-6) Area of inner circle = r2= 22/7 30 = 30 units square Area of...
Read More →If the line y = √3x + k touches the circle
Question: If the line y = 3x + ktouches the circle x2+ y2 = 16, then find the value of k. Solution: Since, the equation of a circle having centre $(h, k)$, having radius as $r$ units, is $(x-h)^{2}+(y-k)^{2}=r^{2}$ $(x-0)^{2}+(y-0)^{2}=4^{2}$ Perpendicular Distance between a point $(0,0) \$ the line $y=\sqrt{3} x+k$ or $\sqrt{3} x-y+k=0$ Perpendicular Distance (Between a point and line) $=$ Whereas the point is $\left(x_{1}, y_{1}\right)$ and the line is expressed as $a x+b y+c=0$ $D=\frac{\{|\s...
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. $9 x^{2}+y^{2}=36$ Solution: Given $9 x^{2}+y^{2}=36$ Divide by 36 to both the sides, we get $\frac{9}{36} x^{2}+\frac{1}{36} y^{2}=1$ $\frac{x^{2}}{4}+\frac{y^{2}}{36}=1$ (i) Since, 4 36 So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (ii) Comparing eq. (i) and (ii), ...
Read More →Find the equation of the circle having (1, –2)
Question: Find the equation of the circle having (1, 2) as its centre and passing through 3x + y = 14, 2x + 5y = 18 Solution: Solving the given equations, 3x + y = 14 .1 2x + 5y = 18 ..2 Multiplying the first equation by 5, we get 15x + 5y = 70.3 2x + 5y = 18..4 Subtract equation 4 from 3 we get 13 x = 52, Therefore x = 4 Substituting x = 4, in equation 1, we get 3 (4) + y = 14 y = 14 12 = 2 So, the point of intersection is (4, 2) Since, the equation of a circle having centre (h, k), having radi...
Read More →If one end of a diameter of the circle
Question: If one end of a diameter of the circle x2+ y2 4x 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter. Solution: Given equation of the circle, x2 4x + y2 6y + 11 = 0 x2 4x + 4 + y2 6y + 9 +11 13 = 0 the above equation can be written as x2 2 (2) x + 22+ y2 2 (3) y + 32+11 13 = 0 on simplifying we get (x 2)2+ (y 3)2= 2 (x 2)2+ (y 3)2=(2)2 Since, the equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2 We have centre = (2, ...
Read More →If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0
Question: If the lines 3x 4y + 4 = 0 and 6x 8y 7 = 0 are tangents to a circle, then find the radius of the circle. Solution: Given both lines are parallel \ tangent to the circle. Then distance between these two lines must be the diameter of the circle. Lines: $3 x-4 y+4=0$ $3 x-4 y-3.5=0$ (Equating the co-efficient of the equation) Distance (between 2 || lines) $=\left(\frac{|c-p|}{\sqrt{a^{2}+b^{2}}}\right)$ for the two given equations, $a x+b y+c=0 \ a x+b y+p=0$ Distance $=\left(\frac{|4-(-3...
Read More →A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,
Question: A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible? Solution: Suppose square of side measuring $x \mathrm{~cm}$ is cut off. Then, the length, breadth and height of the box will be $(45-x),(24-2 x)$ and $x$, respectively. $\Rightarrow$ Volume of the box, $V=(45-2 x)(24-2 x) \mathrm{x}$ $\Righ...
Read More →Find the equation of the circle
Question: Find the equation of the circle which touches x-axis and whose centre is (1, 2). Solution: Since the circle has a centre (1, 2) and also touches x-axis. Radius of the circle is, r = 2 The equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2 So, the equation of the required circle is: (x 1)2+ (y 2)2= 22 ⇒x2 2x + 1 + y2 4y + 4 = 4 ⇒x2+ y2 2x 4y + 1 = 0 The equation of the circle isx2+ y2 2x 4y + 1 = 0....
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. $3 x^{2}+2 y^{2}=18$ Solution: Given: $3 x^{2}+2 y^{2}=18$ Divide by 18 to both the sides, we get $\frac{3}{18} x^{2}+\frac{2}{18} y^{2}=1$ $\frac{x^{2}}{6}+\frac{y^{2}}{9}=1$ (i) Since, $69$ So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (ii) Comparing eq. (i) and (i...
Read More →Show that the point (x, y) given
Question: Show that the point $(x, y)$ given by $x=\frac{2 a t}{1+t^{2}}$ and $y=\frac{a\left(1-t^{2}\right)}{1+t^{2}}$ lies on a circle for all real values of $t$ such that $-1 \leq t \leq 1$ where $a$ is any given real numbers. Solution: Given $x=\frac{2 a t}{1+t^{2}}, y=\frac{a\left(1-t^{2}\right)}{1+t^{2}}$ Squaring both the equations, $\mathrm{x}^{2}=\left[\frac{2 \mathrm{at}}{1+\mathrm{t}^{2}}\right]^{2} \ \mathrm{y}^{2}=\left[\frac{\mathrm{a}\left(1-\mathrm{t}^{2}\right)}{1+\mathrm{t}^{2}...
Read More →A square piece of tin of side 18 cm is to be made
Question: A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume. Solution: Let the side of the square to be cut off bexcm.Then, the length and the breadth of the box will be (18 2x) cm each and height of the box will bexcm. Volume of the box, $V(x)=x(18-2 x)^{2}$ $V^{\prime}(x)=(18-2 ...
Read More →A square piece of tin of side 18 cm is to be made
Question: A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume. Solution: Let the side of the square to be cut off bexcm.Then, the length and the breadth of the box will be (18 2x) cm each and height of the box will bexcm. Volume of the box, $V(x)=x(18-2 x)^{2}$ $V^{\prime}(x)=(18-2 ...
Read More →Two sides of a triangle have lengths 'a' and 'b' and the angle between them is θθ.
Question: Two sides of a triangle have lengths 'a' and 'b' and the angle between them is. What valueofwill maximize the area of the triangle? Find the maximum area of the triangle also. Solution: As, the area of the triangle, $A=\frac{1}{2} a b \sin \theta$ $\Rightarrow A(\theta)=\frac{1}{2} a b \sin \theta$ $\Rightarrow A^{\prime}(\theta)=\frac{1}{2} a b \cos \theta$ For maxima or minima, $A^{\prime}(\theta)=0$ $\Rightarrow \frac{1}{2} a b \cos \theta=0$ $\Rightarrow \cos \theta=0$ $\Rightarrow...
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. $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ Solution: Given: $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ ..........(i) Since, $916$ So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ ...........(ii) Comparing eq. (i) and (ii), we get $a^{2}=16$ and $b^{2}=9$ $\Rightarrow a=\sqrt{16}$ a...
Read More →Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.
Question: Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long. Solution: Let the base of the right angled triangle be $x$ and its height be $y$. Then, $x^{2}+y^{2}=5^{2}$ $\Rightarrow y^{2}=25-x^{2}$ $\Rightarrow y=\sqrt{25-x^{2}}$ As, the area of the triangle, $A=\frac{1}{2} \times x \times y$ $\Rightarrow A(x)=\frac{1}{2} \times \mathrm{x} \times \sqrt{25-\mathrm{x}^{2}}$ $\Rightarrow A(x)=\frac{x \sqrt{25-x^{2}}}{2}$ $\Rightarrow A^{\prime}(x)=\frac{\sqrt{2...
Read More →The distance of the point of intersection
Question: The distance of the point of intersection of the lines 2x 3y + 5 = 0 and 3x + 4y = 0 from the line 5x 2y = 0 is A. $\frac{130}{17 \sqrt{29}}$ B. $\frac{13}{7 \sqrt{29}}$ c. $\frac{130}{7}$ D. None of these Solution: A. $\frac{130}{17 \sqrt{29}}$ Explanation: Giventwo lines are 2x 3y + 5 = 0 (i) And3x + 4y = 0 (ii) Now, point of intersection of these lines can be find out as Multiplying equation (i) by 3, we get 6x 9y + 15 = 0 (iii) Multiplying equation (ii) by 2, we get 6x + 8y = 0 (iv...
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. $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ Solution: Given: $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ (i) Since, $425$ So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (ii) Comparing eq. (i) and (ii), we get $a^{2}=25$ and $b^{2}=4$ $\Rightarrow a=\sqrt{25}$ and $b=\sqrt{4}$ $\Rig...
Read More →Given the sum of the perimeters of a square and a circle,
Question: Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle. Solution: Let the length of a side of the square and radius of the circle be $x$ and $r$, respectively. It is given that the sum of the perimeters of square and circle is constant. $\Rightarrow 4 x+2 \pi r=K$ (Where $K$ is some constant) $\Rightarrow x=\frac{(K-2 \pi r)}{4}$ .....(1) Now, $A=x^{2}+\pi r^{2}$ $\Rightarrow A=\...
Read More →Prove the following
Question: If the line $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$ passes through the points $(2,-3)$ and $(4,-5)$, then $(\mathrm{a}, \mathrm{b})$ is A. (1, 1) B. ( 1, 1) C. (1, 1) D. ( 1, 1) Solution: D. ( 1, 1) Explanation: Given points are $(2,-3)$ and $(4,-5)$ Firstly, we find the equation of line. We know that, Equation of line when two points are given: $y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$ Putting the values, we get $y-(-3)=\frac{-5-(-3)}{4-2}(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. $4 x^{2}+9 y^{2}=1$ Solution: Given: $4 x^{2}+9 y^{2}=1$ $\frac{x^{2}}{\frac{1}{4}}+\frac{y^{2}}{\frac{1}{9}}=1$ (i) Since, $\frac{1}{4}\frac{1}{9}$ 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 $a^{2}=\frac{1}{4}$ and $b^{2}=\...
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