Find the equation of an ellipse, the lengths of whose major and mirror axes
Question: Find the equation of an ellipse, the lengths of whose major and mirror axes are 10 and 8 units respectively. Solution: Let the equation of required ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (A) Given: Length of Major Axis = 10units ...(i) We know that, Length of major axis $=2 \mathrm{a} \ldots$ (ii) $\therefore$ From eq. (i) and (ii), we get 2a = 10 ⇒ a = 5 It is also given that Length of Minor Axis $=8$ units ..(iii) We know that, Length of minor axis $=2 \mathrm{~b} \ld...
Read More →Three consecutive vertices of a parallelogram
Question: Three consecutive vertices of a parallelogram ABCD are A (6, 2, 4), B (2, 4, 8), C (2, 2, 4). Find the coordinates of the fourth vertex. Solution: Given three consecutive vertices of a parallelogram $A B C D$ are $A(6,-2,4), B(2,$, $4,-8), C(-2,2,4)$. Let the forth vertex be $\mathrm{D}(\mathrm{x}, \mathrm{y}, \mathrm{z})$. Midpoint of diagonal $A C=P\left(\frac{6-2}{2}, \frac{-2+2}{2}, \frac{4+4}{2}\right)=P(2,0,4)$ Midpoint of diagonal $\mathrm{BD}=\mathrm{P}\left(\frac{\mathrm{x}+2}...
Read More →Determine the points on the curve
Question: Determine the points on the curve $x 2=4 y$ which are nearest to the point $(0,5)$. Solution: Let the point $(x, y)$ on the curve $x^{2}=4 y$ be nearest to $(0,5)$. Then, $x^{2}=4 y$ $\Rightarrow y=\frac{x^{2}}{4}$ $\ldots(1)$ Also, $d^{2}=(x)^{2}+(y-5)^{2}$ [Using distance formula] Now, $Z=d^{2}=(x)^{2}+(y-5)^{2}$ $\Rightarrow Z=(x)^{2}+\left(\frac{x^{2}}{4}-5\right)^{2}$ $\left[\begin{array}{ll}\text { Using eq. } \text { (1) }\end{array}\right]$ $\Rightarrow Z=x^{2}+\frac{x^{4}}{16}...
Read More →Show that the point A (1, – 1, 3),
Question: Show that the point A (1, 1, 3), B (2, 4, 5) and (5, 13, 11) are collinear. Solution: Given points are $A(1,-1,3), B(2,-4,5)$ and $(5,-13,11)$. To prove collinear, $\mathrm{AB}=\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}}=\sqrt{1+9+4}=\sqrt{14}$ $\mathrm{BC}=\sqrt{(2-5)^{2}+(-4+13)^{2}+(5-11)^{2}}=\sqrt{9+81+36}=3 \sqrt{14}$ $\mathrm{AC}=\sqrt{(1-5)^{2}+(-1+13)^{2}+(3-11)^{2}}=\sqrt{16+144+64}=4 \sqrt{14}$ $\therefore \mathrm{AB}+\mathrm{BC}=\sqrt{14}+3 \sqrt{14}$ $=4 \sqrt{14}$ $=\mathrm{AC}...
Read More →Show that if
Question: Show that if $x^{2}+y^{2}=1$, then the point $\left(x, y \sqrt{1-x^{2}-y^{2}}\right)$ is at a distance 1 unit from the origin. Solution: Given $x^{2}+y^{2}=1 \Rightarrow 1-x^{2}-y^{2}=0$ Distance of the point $\left(x, y \sqrt{1-x^{2}-y^{2}}\right)$ from origin is $=$ $\sqrt{x^{2}+y^{2}+\left(\sqrt{1-x^{2}-y^{2}}\right)^{2}}$ $=\sqrt{x^{2}+y^{2}+\left(1-x^{2}-y^{2}\right)}$ $=\sqrt{x^{2}+y^{2}+1-x^{2}-y^{2}}$ $=\sqrt{1}$ $=1$ unit. When $x=1$ the distance of that point from origin will...
Read More →Find the equation of the ellipse which passes through the point
Question: Find the equation of the ellipse which passes through the point (4, 1) and having its foci at (3, 0). Solution: Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$(i) Given: Coordinates of foci $=(\pm 3,0)$...(ii) We know that, Coordinates of foci $=(\pm C, 0)$...(iii) $\therefore$ From eq. (ii) and (iii), we get c = 3 We know that, $c^{2}=a^{2}-b^{2}$ $\Rightarrow(3)^{2}=a^{2}-b^{2}$ $\Rightarrow 9=a^{2}-b^{2}$ $\Rightarrow b^{2}=a^{2}-9 \ldots$ (iv...
Read More →Show that among all positive numbers
Question: Show that among all positive numbers $x$ and $y$ with $x 2+y 2=r 2$, the sum $x+y$ is largest when $x=y=r \sqrt{2}$. Solution: Here, $x^{2}+y^{2}=r^{2}$ $\Rightarrow y=\sqrt{r^{2}-x^{2}}$ ......(1) Now, $Z=x+y$ $\Rightarrow Z=x+\sqrt{r^{2}-x^{2}}$ [From eq. (1)] $\Rightarrow \frac{d Z}{d x}=1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}$ For maximum or minimum values of $Z$, we must have $\frac{d Z}{d x}=0$ $\Rightarrow 1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}=0$ $\Rightarrow 2 x=2 \sqrt{r^{2}-x^{...
Read More →Show that the maximum volume of the cylinder
Question: Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5 \sqrt{3 \mathrm{~cm}} i s 500 \pi \mathrm{cm}^{3}$. Solution: Let the height, radius of base and volume of a cylinder be $h, r$ and $V$, respectively. Then, $\frac{h^{2}}{4}+r^{2}=R^{2}$ $\Rightarrow h^{2}=4\left(R^{2}-r^{2}\right)$ $\Rightarrow r^{2}=R^{2}-\frac{h^{2}}{4}$ $\cdots(1)$ Now, $V=\pi r^{2} h$ $\Rightarrow V=\pi\left(h R^{2}-\frac{h^{3}}{4}\right)$ $[$ From eq. $(1)]$ $\Rightarrow ...
Read More →Find the equation of the ellipse with eccentricity
Question: Find the equation of the ellipse with eccentricity $\frac{3}{4}$, foci on the $y$-axis, center at the origin and passing through the point (6, 4). Solution: Given that Eccentricity $=\frac{3}{4}$ we know that, Eccentricity, $\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}$ $\Rightarrow \frac{3}{4}=\frac{c}{a}$ $\Rightarrow c=\frac{3}{4} a$ We know that, $c^{2}=a^{2}-b^{2}$ $\Rightarrow\left(\frac{3 a}{4}\right)^{2}=a^{2}-b^{2}$ $\Rightarrow \frac{9 a^{2}}{16}=a^{2}-b^{2}$ $\Rightarrow b^{2}=a...
Read More →A closed cylinder has volume
Question: A closed cylinder has volume $2156 \mathrm{~cm}^{3}$. What will be the radius of its base so that its total surface area is minimum. Solution: Let the height, radius of the base and surface area of the cylinder be $h, r$ and $S$, respectively. Then, Volume $=\pi r^{2} h$ $\Rightarrow 2156=\pi r^{2} h$ $\Rightarrow 2156=\frac{22}{7} r^{2} h$ $\Rightarrow h=\frac{2156 \times 7}{22 r^{2}}$ $\Rightarrow h=\frac{686}{r^{2}}$ ......(1) Surface area $=2 \pi r h+2 \pi r^{2}$ $\Rightarrow S=\fr...
Read More →Show that the height of the cone of maximum volume
Question: Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius $12 \mathrm{~cm}$ is $16 \mathrm{~cm}$. Solution: Let the height, radius of base and volume of the cone be $h, r$ and $V$, respectively. Then, $h=R+\sqrt{R^{2}-r^{2}}$ $\Rightarrow h-R=\sqrt{R^{2}-r^{2}}$ Squaring both the sides, we get $h^{2}+R^{2}-2 h R=R^{2}-r^{2}$ $\Rightarrow r^{2}=2 h R-h^{2}$ ....(1) Now, $V=\frac{1}{3} \pi r^{2} h$ $\Rightarrow V=\frac{\pi}{3}\left(2 h^{2} R-h^{3}\rig...
Read More →Find the equation of the ellipse with center at the origin, the major axis on
Question: Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points (4, 3) and (-1, 4). Solution: Given: Center is at the origin and Major axis is along $x$-axis So, Equation of ellipse is of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (i) Given that ellipse passing through the points $(4,3)$ and $(-1,4)$ So, point $(4,3)$ and $(-1,4)$ will satisfy the eq. (i) Taking point $(4,3)$ where $x=4$ and $y=3$ Putting the values in e...
Read More →Find the distance from the origin
Question: Find the distance from the origin to (6, 6, 7). Solution: The distance from the origin to $(6,6,7)=$ $\sqrt{6^{2}+6^{2}+7^{2}}$ $=\sqrt{36+36+49}$ $=\sqrt{121}$ $=11$ units....
Read More →How far apart are the points (2, 0, 0)
Question: How far apart are the points (2, 0, 0) and (3, 0, 0)? Solution: The points(2, 0, 0) and (3, 0, 0) are at a distance of = |2 (3)| = 5 units....
Read More →Let A, B, C be the feet of perpendiculars
Question: Let A, B, C be the feet of perpendiculars from a point P on the xy, yz and zx planes respectively. Find the coordinates of A, B, C in each of the following where the point P is (i) (3, 4, 5) (ii) (5, 3, 7) (iii) (4, 3, 5). Solution: (i) (3, 4, 5):- A (3, 4, 0), B (0, 4, 5), C (3, 0, 5) (ii) (5, 3, 7):- A (5, 3, 0), B (0, 3, 7), C (5, 0, 7) (iii) (4, 3, 5):- A (4, 3, 0), B (0, 3, 5), C (4, 0, 5)...
Read More →Let A, B, C be the feet of perpendiculars from
Question: Let A, B, C be the feet of perpendiculars from a point P on the x, y, z-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is : (i) A = (3, 4, 2) (ii) (5, 3, 7) (iii) (4, 3, 5) Solution: (i) (3, 4, 2):- A (3, 0, 0), B (0, 4, 0), C (0, 0, 2) (ii) (5, 3, 7):- A (5, 0, 0), B (0, 3, 0), C (0, 0, 7) (iii) (4, 3, 5):- A (4, 0, 0), B (0, 3, 0), C (0, 0, 5)...
Read More →Find the dimensions of the rectangle of perimeter 36 cm
Question: Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides. Solution: Let $l, b$ and $V$ be the length, breadth and volume of the rectangle, respectively. Then, $2(l+b)=36$ $\Rightarrow l=18-b$ .....(1) Volume of the cylinder when revolved about the breadth, $V=\pi l^{2} b$ $\Rightarrow V=\pi(18-b)^{2} b$ $[$ From eq. $(1)]$ $\Rightarrow V=\pi\left(324 b+b^{3}-36 b^{2}\right)$ $\Rightarrow \frac{d V}{d...
Read More →Name the octant in which each of the following points lies.
Question: Name the octant in which each of the following points lies. (i) (1, 2, 3), (ii) (4, 2, 3), (iii) (4, 2, 5) (iv) (4, 2, 5) (v) ( 4, 2, 5) (vi) (3, 1, 6) (vii) (2, 4, 7) (viii) ( 4, 2, 5). Solution: (i) (1, 2, 3):- 1stOctant, (ii) (4, 2, 3):- 4thOctant, (iii) (4, 2, 5):- 8thOctant, (iv) (4, 2, 5):- 5thOctant, (v) ( 4, 2, 5):- 2ndOctant, (vi) (3, 1, 6):- 3rdOctant, (vii) (2, 4, 7):- 8thOctant, (viii) ( 4, 2, 5):- 6th Octant....
Read More →Find the equation of the ellipse whose foci are at (0, ±4) and
Question: Find the equation of the ellipse whose foci are at $(0, \pm 4)$ and $\mathrm{e}=\frac{4}{5} .$ Solution: Given: Coordinates of foci $=(0, \pm 4) \ldots$ (i) We know that, Coordinates of foci $=(0, \pm c) \ldots$ (ii) The coordinates of the foci are $(0, \pm 4)$. This means that the major and minor axes are along $y$ and $x$ axes respectively. $\therefore$ From eq. (i) and (ii), we get c = 4 It is also given that Eccentricity $=\frac{4}{5}$ we know that, Eccentricity, $e=\frac{c}{a}$ $\...
Read More →Find the equation of each of the following parabolas
Question: Find the equation of each of the following parabolas (a) Directrix x = 0, focus at (6, 0) (b) Vertex at (0, 4), focus at (0, 2) (c) Focus at (1, 2), directrix x 2y + 3 = 0 Solution: (a) The distance of any point on the parabola from its focus and its directrix is same. Given that, directrix, x = 0 and focus = (6, 0) If a parabola has a vertical axis, the standard form of the equation of the parabola is (x h)2= 4p (y k), where p 0. The vertex of this parabola is at (h, k). The focus is ...
Read More →Find the equation of the ellipse whose foci are at
Question: Find the equation of the ellipse whose foci are at (1, 0) and $e=\frac{1}{2}$ Solution: Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Given: Coordinates of foci $=(\pm 1,0) \ldots(\mathrm{i})$ We know that, Coordinates of foci $=(\pm c, 0) \ldots$ (ii) $\therefore$ From eq. (i) and (ii), we get c = 1 It is also given that Eccentricity $=\frac{1}{2}$ we know that, Eccentricity, $e=\frac{c}{a}$ $\Rightarrow \frac{1}{2}=\frac{1}{a}[\because c=1]$ ...
Read More →Prove that the least perimeter of an isosceles triangle i
Question: Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6 \sqrt{3} \mathrm{r}$. Solution: Let $\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}=x$ and a circle with centre $\mathrm{O}$ and radius $r$ is inscribed in the triangle. $\mathrm{O}, \mathrm{A}$ and $\mathrm{O}, \mathrm{E}$ and $\mathrm{O}, \mathrm{D}$ are joined. From $\Delta \mathrm{ABF}$ $\mathrm{AF}^{2}+\mathrm{BF}^{2}=\mathrm{AB}^{2}$ $\Rightarrow(3 ...
Read More →Find the equation of the ellipse whose foci are
Question: Find the equation of the ellipse whose foci are (2, 0) and the eccentricity is $\frac{1}{2}$ Solution: Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Given: Coordinates of foci $=(\pm 2,0)$...(iii) We know that, Coordinates of foci $=(\pm c, 0)$...(iv) $\therefore$ From eq. (iii) and (iv), we get $c=2$ It is also given that Eccentricity $=\frac{1}{2}$ we know that, Eccentricity, e $=\frac{c}{a}$ $\Rightarrow \frac{1}{2}=\frac{2}{a}[\because c=2]...
Read More →Find the equation of a circle passing through
Question: Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x 1. Solution: Since, the equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2Centre lies on the line i.e., y = x 1, Co Ordinates are (h, k) = (h, h 1) (x h)2+ (y k)2= r2 (7 h)2+ (3 (h 1))2= 32 49 + h2 14h + (3 h +1)2= 9 On rearranging we get h2 14h + 49 +16 +h2 8h 9 = 0 2h2 22h + 56 = 0 h2 11h + 28 = 0 h2 4h 7h + 28 = 0 h (h 4...
Read More →An isosceles triangle of vertical angle
Question: An isosceles triangle of vertical angle $2 \theta$ is inscribed in a circle of radius $a$. Show that the area of the triangle is maximum when $\theta=\frac{\pi}{6}$. Solution: Let $\mathrm{ABC}$ be an isosceles triangle inscribed in the circle with radius a such that $\mathrm{AB}=\mathrm{AC}$. $\mathrm{AD}=\mathrm{AO}+\mathrm{OD}=a+a \cos 2 \theta=a(1+\cos 2 \theta)$ and $\mathrm{BC}=2 \mathrm{BD}=2 a \sin 2 \theta$ As, area of the triangle $\mathrm{AC}, A=\frac{1}{2} \mathrm{BC} \time...
Read More →