Exercise 5.4 - Chapter 5 Two Dimensional Analytical Geometry–II 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $5.4$
Question 1.
Find the equations of the two tangents that can be drawn from $(5,2)$ to the ellipse $2 x^2+7 y^2=14$.
Solution:
$
\begin{aligned}
& 2 \mathrm{x}^2+7 \mathrm{y}^2=14 \\
& (\div \text { by } 14) \Rightarrow \frac{x^2}{7}+\frac{y^2}{2}=1
\end{aligned}
$
comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
we get $\mathrm{a}^2=7$ and $\mathrm{b}^2=2$
The equation of tangent to the above ellipse will be of the form
$
\mathrm{y}=\mathrm{mx}+\sqrt{a^2 m^2+b^2} \Rightarrow \mathrm{y}=\mathrm{mx}+\sqrt{7 m^2+2}
$
Here the tangents are drawn from the point $(5,2)$
$
\Rightarrow 2=5 \mathrm{~m}+\sqrt{7 m^2+2} \Rightarrow 2-5 \mathrm{~m}=\sqrt{7 m^2+2}
$
Squaring on both sides we get
$
\begin{aligned}
& (2-5 \mathrm{~m})^2=7 \mathrm{~m}^2+2 \\
& 25 \mathrm{~m}^2+4-20 \mathrm{~m}-7 \mathrm{~m}^2-2=0 \\
& 18 \mathrm{~m}^2-20 \mathrm{~m}+2=0 \\
& (\div \text { by } 2) \Rightarrow 9 \mathrm{~m}^2-10 \mathrm{~m}+1=0 \\
& (9 \mathrm{~m}-1)(\mathrm{m}-1)=0 \\
& \mathrm{~m}=1 \text { (or) } \mathrm{m}=1 / 9
\end{aligned}
$
When $\mathrm{m}=1$, the equation of tangent is
$
y=x+3 \text { or } x-y+3=0
$
When $\mathrm{m}=\frac{1}{9}$ the equation of tangent is 9
$
\begin{aligned}
& \mathrm{y}==\frac{x}{9}+\sqrt{\frac{7}{81}+2} \text { (i.e.) } \mathrm{y}=\frac{x}{9}+\frac{13}{9} \\
& 9 \mathrm{y}=\mathrm{x}+13 \text { or } \mathrm{x}-9 \mathrm{y}+13=0
\end{aligned}
$
Question 2.
Find the equations of tangents to the hyperbola $\frac{x^2}{16}-\frac{y^2}{64}=1$ which are parallel to $10 \mathrm{x}-3 \mathrm{y}+9=0$.
Solution:
$
\frac{x^2}{16}-\frac{y^2}{64}=1
$
Here $a^2=16$ and $b^2=64$
The equation of tangents will be of the form $\mathrm{y}=\mathrm{mx} \pm \sqrt{a^2 m^2-b^2}$
$
\text { (i.e.,) } y=m x \pm \sqrt{16 m^2-64}
$
Where ' $m$ ' is the slope of the tangent.
Here we are given that the tangents are parallel to $10 x-3 y+9=0$

So slope of tangents will be equatl to slope of the given line
(ie.,)
$
m=-\left(\frac{10}{-3}\right)=\frac{10}{3}
$
So equations of tangents will be
$
\begin{aligned}
y & =\frac{10}{3} x \pm \sqrt{16\left(\frac{10}{3}\right)^2-64}=\frac{10 x}{3} \pm \sqrt{16\left(\frac{100}{9}\right)-64} \\
y & =\frac{10 x}{3} \pm \sqrt{\frac{1600-576}{9}} \Rightarrow y=\frac{10 x}{3} \pm \frac{32}{3} \\
\Rightarrow \quad 3 y & =10 x \pm 32 \text { (or) } 10 x-3 y \pm 32=0 .
\end{aligned}
$

Question 3.
Show that the line $x-y+4=0$ is a tangent to the ellipse $x^2+3 y^2=12$. Also find the coordinates of the point of contact.
Solution:
The given ellipse is $\mathrm{x}^2+3 \mathrm{y}^2=12$
$(\div$ by 12$) \Rightarrow \frac{x^2}{12}+\frac{y^2}{4}=1$
(ie.,) Here $\mathrm{a}^2=12$ and $\mathrm{b}^2=4$
The given line is $x-y+4=0$
(ie.,) $y=x+4$
Comparing this line with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
We get $\mathrm{m}=1$ and $\mathrm{c}=4$
The condition for the line $y=m x+c$
To be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2$
LHS $=c^2=4^2=16$
RHS: $a^2 \mathrm{~m}^2+b^2=12(1)^2+4=16$
LHS $=$ RHS The given line is a tangent to the ellipse. Also the point of contact is $\left(\frac{-a^2 m}{c}, \frac{b^2}{c}\right)=\left[-\left(\frac{12(1)}{4}\right), \frac{4}{4}\right] \quad$ (i.e.,) $(-3,1)$

Question 4.
Find the equation of the tangent to the parabola $y^2=16 x$ perpendicular to $2 x+2 y+3=0$

Solution:
$
y^2=16 x
$
Comparing this equation with $\mathrm{y}^2=4 \mathrm{ax}$
we get $4 \mathrm{a}=16 \Rightarrow \mathrm{a}=4$
The equation of tangent to the parabola $y^2=16 x$ will be of the form
$
\begin{aligned}
y & =m x+\frac{a}{m}(\text { Here } a=4) \\
\Rightarrow \quad y & =m x+\frac{4}{m}
\end{aligned}
$
Given that the tangent is perpendicular to $2 x+2 y+3=0$
$\left[\begin{array}{l}\text { So slope of } 2 x+2 y+3=0 \text { is }-\left(\frac{2}{2}\right)=-1=m_1 \\ \Rightarrow \text { slope of its perpendicular }=\frac{-1}{m_1}=\frac{-1}{-1}=1 \\ \text { (ie.,) } m=1\end{array}\right]$
So $\mathrm{m}=1$
$\Rightarrow$ The equation of tangent will be $\mathrm{y}=1(\mathrm{x})+\frac{4}{1}$ (i.e.) $y=\mathrm{x}+4$
(or) $x-y+4=0$

Question 5.
Find the equation of the tangent at $t=2$ to the parabola $y^2=8 x$. (Hint: use parametric form)

Solution:

$\mathrm{y}^2=8 \mathrm{x}$
Comparing this equation with $y^2=4 a x$ we get $4 a=8 \Rightarrow a=2$
Now, the parametric form for $\mathrm{y}^2=4 a x$ is $\mathrm{x}=\mathrm{at}^2, \mathrm{y}=2 \mathrm{at}$ Here $\mathrm{a}=2$ and $\mathrm{t}=2$ $\Rightarrow \mathrm{x}=2(2)^2=8$ and $\mathrm{y}=2(2)(2)=8$
So the point is $(8,8)$
Now eqution of tangent to $\mathrm{y}^2=4$ ax at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{yy}_1=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)$
Here $\left(\mathrm{x}_1, \mathrm{y}_1\right)=(8,8)$ and $\mathrm{a}=2$
So equation of tangent is $y(8)=2(2)(x+8)$
(ie.,) $8 y=4(x+8)$
$(\div$ by 4$) \Rightarrow 2 \mathrm{y}=\mathrm{x}+8 \Rightarrow \mathrm{x}-2 \mathrm{y}+8=0$
Aliter
The equation of tangent to the parabola $y^2=4 a x$ at ' $t$ ' is
$\mathrm{yt}=\mathrm{x}+\mathrm{at}^2$
Here $t=2$ and $a=2$
So equation of tangent is
(i.e.,) $y(2)=x+2(2)^2$
$2 \mathrm{y}=\mathrm{x}+8 \Rightarrow \mathrm{x}-2 \mathrm{y}+8=0$

Question 6.
Find the equations of the tangent and normal to hyperbola $12 x^2-9 y^2=108$ at $\theta=\frac{\pi}{3}$.
(Hint: Use parametric form)
Solution:
$
12 x^2-9 y^2=108
$
$
\begin{aligned}
\left(\div \text { by 108) we get } \frac{12 x^2}{108}-\frac{9 y^2}{108}\right. & =1 \\
\frac{x^2}{9}-\frac{y^2}{12} & =1
\end{aligned}
$
Comparing this equation with the general form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
we get $a^2=9$ and $b^2=12$
The parametric form for the hyperbola
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is }(a \sec \theta, b \tan \theta)
$
Here $a=3$ and $b=\sqrt{12}=2 \sqrt{3}$
Also $\theta=\frac{\pi}{3}$
So $a \sec \theta=3 \times 2=6=x_1$ and $b \tan \theta=2 \sqrt{3}(\sqrt{3})=6=y_1$
So $\left(x_1, y_1\right)=(6,6)$
Now equation of tangent to $\frac{x^2}{9}-\frac{y^2}{12}=1$ at $\left(x_1, y_1\right)$ is $\frac{x x_1}{9}-\frac{y y_1}{12}=1$ Here $\left(x_1, y_1\right)=(6,6)$
So equation of tangent is
$
\frac{x(6)}{9}-\frac{y(6)}{12}=1 \Rightarrow \frac{2 x}{3}-\frac{y}{2}=1 \Rightarrow 4 x-3 y-6=0
$
Normal is a line perpendicular to tangent
So equation of normal will be of the form $3 x+4 y+k=0$
The normal is drawn at $(6,6)$
$
\Rightarrow 18+24+\mathrm{k}=0 \Rightarrow \mathrm{k}=-42
$

So equation of normal is $3 x+4 y-42=0$
Aliter
The equation of tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $\theta$ is $\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$
Here $a=3, b=2 \sqrt{3}$ and $\theta=\frac{\pi}{3}$
So $\sec \theta=\sec \frac{\pi}{3}=2$ and $\tan \theta=\tan \frac{\pi}{3}=\sqrt{3}$
So equation of tengent will be
$
\begin{aligned}
\frac{x}{3}(2)-\frac{y}{2 \sqrt{3}}(\sqrt{3}) & =1 \\
\frac{2 x}{3}-\frac{y}{2} & =1 \Rightarrow 4 x-3 y-6=0
\end{aligned}
$
(ie.,)
Equation of normal in parametric form is $\frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2+b^2$
$
\text { (ie.,) } \begin{aligned}
\frac{3(x)}{2}+\frac{2 \sqrt{3} y}{\sqrt{3}} & =9+12 \\
\frac{3 x}{2}+2 y & =21 \Rightarrow 3 x+4 y=42 \\
\Rightarrow \quad 3 x+4 y-42 & =0
\end{aligned}
$
Question 7.
Prove that the point of intersection of the tangents at ' $t_1$ ' and $t_2$ ' on the parabola $y^2=4 a x$ is $\left[a t_1 t_2, a\left(t_1+\right.\right.$ $\left.\left.\mathrm{t}_2\right)\right]$.
Solution:
The equation of tangent to $y^2=4 a x$ at ' $t$ ' is given by $y t=x+a t^2$
So equation of tangent at ' $t_1$ ' is $\mathrm{yt}_1=\mathrm{x}+\mathrm{at}_1{ }^2$
and equation of tangent at ' $\mathrm{t}_2$ ' is $\mathrm{yt}_2=\mathrm{x}+\mathrm{at}_2{ }^2$

To find the point of intersection we have to solve the two equations
(1) $\Rightarrow \quad y t_1=x+a t_1^2$.
(2) $\Rightarrow \quad y t_2=x+a t_2^2$
$
\begin{aligned}
(1)-(2) \Rightarrow \quad y\left(t_1-t_2\right) & =a\left(t_1^2-t_2^2\right) \\
\Rightarrow \quad y=\frac{a\left(t_1+t_2\right)\left(t_1-t_2\right)}{\left(t_1-t_2\right)} & =a\left(t_1+t_2\right) \\
y & =a\left(t_1+t_2\right)
\end{aligned}
$
Substituting $y$ value in (1) we get
$
\begin{aligned}
& a\left(t_1+t_2\right)\left(t_1\right) & =x+a t_1^2 \\
(i e .,) & a t_1^2+a t_1 t_2 & =x+a t_1^2 \\
\Rightarrow & x & =a t_1 t_2
\end{aligned}
$
So the point of intersection is $\left[a t_1 t_2, a\left(t_1+t_2\right)\right]$
Question 8.
If the normal at the point ' $t_1$ ' on the parabola $y^2=4 a x$ meets the parabola again at the point ${ }^{\prime} t_2$ ', then prove that $\mathrm{t}_2=-\left(t_1+\frac{2}{t_1}\right)$
Solution:
Equation of normal to $y^2=4$ at' $t$ ' is $y+x t=2 a t+a t^3$.
So equation of normal at ' $t_1$ ' is $y+x t_1=2 a t_1+a_1{ }^3$
The normal meets the parabola $y^2=4 a x$ at ' $t_2$ ' (ie.,) at $\left(a t_2{ }^2, 2 a t_2\right.$ )
$
\begin{aligned}
& \Rightarrow 2 \mathrm{at}_2+\mathrm{at}_1 \mathrm{t}_2^2=2 \mathrm{at}_1+\mathrm{at}_1{ }^3 \\
& \text { So } 2 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right)=\mathrm{at}_1{ }^3-\mathrm{at}_1 \mathrm{t}_2{ }^2 \\
& \Rightarrow 2 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right)=\mathrm{at}_1\left(\mathrm{t}_1{ }^2-\mathrm{t}_2{ }^2\right) \\
& \Rightarrow 2\left(\mathrm{t}_2-\mathrm{t}_1\right)=\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)\left(\mathrm{t}_1-\mathrm{t}_2\right) \\
& \Rightarrow 2=-\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right) \\
& \Rightarrow \mathrm{t}_1+\mathrm{t}_2=\frac{-2}{t_1} \\
& \Rightarrow \mathrm{t}_2=-t_1-\frac{2}{t_1}=-\left(t_1+\frac{2}{t_1}\right)
\end{aligned}
$