Maths Solve the Following Question.(3 Marks)

Comparing this equation with $\frac{x^2}{a^2}-\frac{y^2}{h^2}=1$, we get

$\begin{aligned} & a^2=25 \text { and } b^2=16 \\ & a=5 \text { and } b=4\end{aligned}$

Equation of tangent at P(θ) is

$\frac{x \sec \theta}{ a }-\frac{y \tan \theta}{ b }=1$

The equation of tangent at P(30°) is

$\begin{aligned} & \frac{x \sec 30^{\circ}}{5}-\frac{y \tan 30^{\circ}}{4}=1 \\ & \frac{2 x}{5 \sqrt{3}}-\frac{y}{4 \sqrt{3}}=1 \\ & 8 x-5 y=20 \sqrt{3}\end{aligned}$

x = 3 sec θ, y = 5 tan θ

$\begin{aligned} & \text { Since } \sec ^2 \theta-\tan ^2 \theta=1 \\ & \frac{x^2}{9}-\frac{y^2}{25}=1\end{aligned}$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$, we get

$\begin{aligned} & a^2=9 \text { and } b^2=25 \\ & a=3 \text { and } b=5\end{aligned}$

Equation of tangent at P(θ) is

$\frac{x \sec \theta}{ a }-\frac{y \tan \theta}{ b }=1$

∴ Equation of tangent at P(π/3) is

$\begin{aligned} & \frac{x \sec \left(\frac{\pi}{3}\right)}{3}-\frac{y \tan \left(\frac{\pi}{3}\right)}{5}=1 \\ & \frac{2 x}{3}-\frac{\sqrt{3} y}{5}=1 \\ & 10 x-3 \sqrt{3} y=15\end{aligned}$

$\frac{x^2}{\left(\frac{51}{7}\right)}-\frac{y^2}{17}=1$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$,

we get

$a^2=\frac{51}{7}$ and $b^2=17$

Equation of the tangent to the hyperbola

$\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{x x_1}{ a ^2}-\frac{y y_1}{ b ^2}=1$

Equation of the tangent at $(-3,-2)$ is

$\begin{aligned} & \frac{-3 x}{\left(\frac{51}{7}\right)}+\frac{2 y}{17}=1 \\ & \frac{-7 x}{17}+\frac{2 y}{17}=1 \\ & 7 x-2 y+17=0\end{aligned}$

Length of conjugate axis = 2b

Given, length of conjugate axis = 3

∴ 2b = 3

$\begin{aligned} & \therefore b=\frac{3}{2} \\ & \therefore b^2=\frac{9}{4}\end{aligned}$

Distance between foci = 2ae

Given, distance between foci = 5

∴ 2ae = 5

$\begin{aligned} & \therefore ae =\frac{5}{2} \\ & \therefore a ^2 e ^2=\frac{25}{4}\end{aligned}$

$\begin{aligned} & \text { Now, } b^2=a^2\left(e^2-1\right) \\ & \therefore b^2=a^2 e^2-a^2 \\ & \therefore \frac{9}{4}=\frac{25}{4}-a^2 \\ & \therefore a^2=\frac{25}{4}-\frac{9}{4} \\ & \therefore a^2=4\end{aligned}$

$\therefore$ The required equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{\left(\frac{9}{4}\right)}=1$

i.e. $\frac{x^2}{4}-\frac{4 y^2}{9}=1$

Given, eccentricity $( e )=\frac{3}{2}$

Distance between foci = 2ae

Given, distance between foci = 12

∴ 2ae = 12

$\begin{aligned} & \therefore 2 a \left(\frac{3}{2}\right)=12 \\ & \therefore 3 a =12 \\ & \therefore a =4 \\ & \therefore a ^2=16\end{aligned}$

$\begin{aligned} & \text { Now, } b ^2= a ^2\left( e ^2-1\right) \\ & \therefore b ^2=\left[\left(\frac{3}{2}\right)^2-1\right] \\ & \therefore b ^2=16\left(\frac{9}{4}-1\right) \\ & \therefore b^2=16\left(\frac{5}{4}\right) \\ & \therefore b^2=20\end{aligned}$

$\therefore$ The required equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{30}=1$

Length of conjugate axis = 2b

Given, length of conjugate axis = 5

2b = 5

$\begin{aligned} & b=\frac{5}{2} \\ & b^2=\frac{25}{4}\end{aligned}$

Distance between foci = 2ae

Given, distance between foci = 13

2ae = 13

$\begin{aligned} & \text { ae }=\frac{13}{2} \\ & a^2 e ^2=\frac{169}{4} \\ & \text { Now, } b^2=a^2\left(e^2-1\right) \\ & b^2=a^2 e^2-a^2 \\ & \frac{25}{4}=\frac{169}{4}-a^2 \\ & a^2=\frac{169}{4}-\frac{25}{4}=36\end{aligned}$

$\therefore$ The required equation of hyperbola is $\frac{x^2}{36}-\frac{y^2}{\frac{25}{4}}=1$

i.e., $\frac{x^2}{36}-\frac{4 y^2}{25}=1$

The ellipse passes through the points (-3, 1) and (2, -2).

∴ Substituting x = -3 and y = 1 in equation of ellipse, we get

$\begin{aligned} & \frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1 \\ & \therefore \frac{9}{a^2}+\frac{1}{b^2}=1\end{aligned}$

Substituting x = 2 and y = -2 in equation of ellipse, we get

$\begin{aligned} & \frac{2^2}{a^2}+\frac{(-2)^2}{b^2}=1 \\ & \therefore \frac{4}{a^2}+\frac{4}{b^2}=1 \ldots \ldots \text {.iii) } \\ & \text { Let } \frac{1}{a^2}= A \text { and } \frac{1}{b^2}= B \end{aligned}$

∴ Equations (i) and (ii) become

9A + B = 1 ..…(iii)

4A + 4B = 1 …..(iv)

Multiplying (iii) by 4, we get

36A + 4B = 4 …..(v)

Subtracting (iv) from (v), we get

32A = 3

$\therefore A =\frac{3}{32}$

Substituting $A=\frac{3}{32}$ in (iv), we get

$\begin{aligned} & 4\left(\frac{3}{32}\right)+4 B=1 \\ & \therefore \frac{3}{8}+4 B=1 \\ & \therefore 4 B=1-\frac{3}{8} \\ & \therefore 4 B=\frac{5}{8} \\ & \therefore B=\frac{5}{32}\end{aligned}$

Since $\frac{1}{a^2}=A$ and $\frac{1}{b^2}=B$

$\begin{aligned} & \frac{1}{a^2}=\frac{3}{32} \text { and } \frac{1}{b^2}=\frac{5}{32} \\ & \therefore a^2=\frac{32}{3} \text { and } b^2=\frac{32}{5}\end{aligned}$

∴ The required equation of ellipse is

$\begin{aligned} & \frac{x^2}{\left(\frac{32}{3}\right)}+\frac{y^2}{\left(\frac{32}{5}\right)} \\ & \text { i.e., } 3 x^2+5 y^2=32\end{aligned}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Then length of major axis = 2a = 10

∴ a = 5

Also, distance between foci= 2ae = 8

∴ 2 × 5 × e = 8

$\begin{aligned} & \therefore e=\frac{4}{5} \\ & \therefore b^2=a^2\left(1-e^2\right) \\ & =25\left(1-\frac{6}{25}\right) \\ & =9\end{aligned}$

$\therefore$ from (1), the equation of the required ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$

Equation of axis is $\mathrm{X}=0$ that is $x+2=0$
Equation of diretrix is $\mathrm{Y}+\mathrm{b}=0$ that is $y+3-1=0$ that is $y+2=0$
Equation of tangent at vertex is $\mathrm{Y}=0$ that is $y+3=0$

$\frac{x^2}{144}-\frac{y^2}{25}=1$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$,

we get

$\begin{array}{ll} & a^2=144 \text { and } b^2=25 \\ \therefore \quad & a=12 \text { and } b=5\end{array}$

Equation of the tangent to the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{h^2}=1$ at $P(\theta)$ is

$\frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1$

Eccentric angle $(\theta)=\frac{\pi}{3}$

$\therefore \quad$ Equation of the tangent at $P\left(\frac{\pi}{3}\right)$ is

$\begin{aligned} & \frac{x \sec \frac{\pi}{3}}{12}-\frac{y \tan \frac{\pi}{3}}{5}=1 \\ \therefore \quad & \frac{2 x}{12}-\frac{\sqrt{3} y}{5}=1 \\ \therefore \quad & \frac{x}{6}-\frac{\sqrt{3} y}{5}=1 \\ \therefore \quad & 5 x-6 \sqrt{3} y=30\end{aligned}$

$\therefore \quad \frac{x^2}{\left(\frac{4}{3}\right)}-\frac{y^2}{4}=1$

Comparing this equation with $\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$,

we get

$a^2=\frac{4}{3}$ and $b^2=4$

Equation of the tangent to the hvperbola

$\frac{x^2}{ a ^2}-\frac{y^2}{ b ^2}=1$ at $\left(x_1, y_1\right)$ is

$\frac{x x_1}{ a ^2}-\frac{y y_1}{ b ^2}=1$

$\therefore \quad$ Equation of the tangent at $(2,2 \sqrt{2})$ is

$\begin{array}{ll} & \frac{2 x}{\left(\frac{4}{3}\right)}-\frac{2 \sqrt{2} y}{4}=1 \\ \therefore \quad & \frac{3 x}{2}-\frac{\sqrt{2} y}{2}=1 \\ \therefore \quad & 3 x-\sqrt{2} y=2\end{array}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Co-ordinates of vertices are (±a, 0).

Given that, co-ordinates of vertices are (±7, 0)

∴ a = 7

Endpoints of the conjugate axis are (0, b) and (0, -b).

Given, the endpoints of the conjugate axis are (0, ±3).

∴ b = 3

The required equation of hyperbola is $\frac{x^2}{7^2}-\frac{y^2}{3^2}=1$

i.e., $\frac{x^2}{49}-\frac{y^2}{9}=1$

$\frac{x^2}{\left(\frac{4}{3}\right)}-\frac{y^2}{4}=1$

Comparing this equation with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we get

$\begin{aligned} a^2 & =\frac{4}{3} \text { and } b^2=4 \\ a & =\frac{2}{\sqrt{3}} \text { and } b=2\end{aligned}$

Length of transverse axis $=2 a=2\left(\frac{2}{\sqrt{3}}\right)=\frac{4}{\sqrt{3}}$

Length of conjugate axis $=2 b=2(2)=4$

We know that

$e =\frac{\sqrt{ a ^2+ b ^2}}{ a }=\frac{\sqrt{\frac{4}{3}+4}}{\left(\frac{2}{\sqrt{3}}\right)}=\frac{\sqrt{\frac{16}{3}}}{\left(\frac{2}{\sqrt{3}}\right)}=\frac{4}{2}=2$

Co-ordinates of foci are $S ( ae , 0)$ and

$S^{\prime}(-a e, 0)$

i.e., $S\left(\frac{2}{\sqrt{3}}(2), 0\right)$ and $S^{\prime}\left(-\frac{2}{\sqrt{3}}(2), 0\right)$,

i.e., $S\left(\frac{4}{\sqrt{3}}, 0\right)$ and $S^{\prime}\left(-\frac{4}{\sqrt{3}}, 0\right)$

Equations of the directrices are $x= \pm \frac{ a }{ e }$.

$x= \pm \frac{\left(\frac{2}{\sqrt{3}}\right)}{2}$

$x= \pm \frac{1}{\sqrt{3}}$

Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(4)}{\left(\frac{2}{\sqrt{3}}\right)}=4 \sqrt{3}$

$\frac{x^2}{3}+\frac{y^2}{6}=1$

Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get

$a^2=3$ and $b^2=6$

Equations of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having slope $\mathrm{m}$ are

$y=m x \pm \sqrt{a^2 m^2+b^2}$

Since (2, 1) lies on both the tangents,

$\begin{aligned} & 1=2 m \pm \sqrt{3 m^2+6} \\ & 1-2 m= \pm \sqrt{3 m^2+6}\end{aligned}$

Squaring both the sides, we get

$\begin{aligned} & 1-4 m+4 m^2=3 m^2+6 \\ & m^2-4 m-5=0 \\ & (m-5)(m+1)=0 \\ & m=5 \text { or } m=-1\end{aligned}$

These are the slopes of the required tangents.

By slope point form $y-y_1=m\left(x-x_1\right)$,

the equations of the tangents are

y – 1 = 5(x – 2) and y – 1 = -1(x – 2)

y – 1 = 5x – 10 and y – 1 = -x + 2

5x – y – 9 = 0 and x + y – 3 = 0

$\frac{x^2}{7}+\frac{y^2}{4}=1$

Comparing this equation with $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$, we get

$a^2=7$ and $b^2=4$

Equations of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ having slope $m$ are

$y=m x \pm \sqrt{a^2 m^2+b^2}$

Since (3, -2) lies on both the tangents,

$\begin{aligned} & -2=3 m \pm \sqrt{7 m^2+4} \\ & -2-3 m= \pm \sqrt{7 m^2+4}\end{aligned}$

Squaring both the sides, we get

$\begin{aligned} & 9 m^2+12 m+4=7 m^2+4 \\ & 2 m^2+12 m=0 \\ & 2 m(m+6)=0 \\ & m=0 \text { or } m=-6\end{aligned}$

These are the slopes of the required tangents.

By slope point form y – y1 = m(x – x1),

the equations of the tangents are

y + 2 = 0(x – 3) and y + 2 = -6(x – 3)

y + 2 = 0 and y + 2 = -6x + 18

y + 2 = 0 and 6x + y – 16 = 0