Exercise 5.4-Additional Problems - Chapter 5 - Two Dimensional Analytical Geometry – II - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Find the equations of the tangent and normal to the parabolas : $x^2+2 x-4 y+4=0$ at $(0,1)$
Solution:
Equation of the tangent at $\left(x_1, y_1\right)$ is
$
x x_1+\frac{2\left(x+x_1\right)}{2}-\frac{4\left(y+y_1\right)}{2}=0 ; \text { Here, }\left(x_1, y_1\right)=(0,1)
$
$\therefore$ Equation of the tangent is
$
\begin{aligned}
& x(0)+\frac{2(x+0)}{2}-\frac{4(y+1)}{2}+4=0 \text {; i.e., } x-\frac{4(y+1)}{2}+4=0 \\
& \frac{2 x-4(y+1)+8}{2}=0 \Rightarrow 2 x-4 y-4+8=0 \\
& 2 x-4 y+4=0 \\
& (\div \text { by } 2) \quad \Rightarrow \quad x-2 y+2=0 \\
& \text { Slope of the tangent }=-\left(\frac{1}{-2}\right)=\frac{1}{2}=m \\
& \therefore \text { Slope of the normal }=\frac{-1}{m}=-2 \\
& \left(x_1, y_1\right)=(0,1) \\
&
\end{aligned}
$
Equation of the normal $y-1=-2(x-0) \Rightarrow y-1=-2 x$
$
\therefore 2 x+y-1=0
$

Question 2.
Find the equations of the tangent and normal to the parabola $y^2=8 x$ at $t=\frac{1}{2}$

Solution:

Equation of the parabola is $y^2=8 x$. Here $4 a=8 \Rightarrow a=2$
$' t_1$ ' represents the point $\left(a t^2, 2 a t\right)=\left(2 t^2, 4 t\right)$; Here, $t=\frac{1}{2}$
$\therefore \quad$ The point is $2\left(\frac{1}{4}\right), 4\left(\frac{1}{2}\right)=\left(\frac{1}{2}, 2\right)$; i.e., $\left(x_1, y_1\right)=\left(\frac{1}{2}, 2\right)$
So equation of the tangent to $y^2=8 x$ at $\left(\frac{1}{2}, 2\right)$ is $y\left(y_1\right)=\frac{8\left(x+x_1\right)}{2}$
$
\begin{aligned}
\text { i.e., } y(2) & =\frac{8\left(x+\frac{1}{2}\right)}{2} \Rightarrow 4 y=\frac{8(2 x+1)}{2} \\
8 y & =16 x+8 \Rightarrow 16 x-8 y+8=0 \\
(\div \text { by } 8) \Rightarrow 2 x-y+1 & =0 \\
\text { Slope of the tangent } & =-\left(\frac{2}{-1}\right)=2=m \\
\therefore \text { Slope of the normal } & =\frac{-1}{m}=\frac{-1}{2} \\
\left(x_1, y_1\right) & =\left(\frac{1}{2}, 2\right)
\end{aligned}
$
Equation of the normal $y-2=\frac{-1}{2}\left(x-\frac{1}{2}\right)$
$
\begin{aligned}
2(y-2) & =-\frac{2 x-1}{2} \\
4(y-2) & =-(2 x-1) \\
4 y-8 & =-2 x+1 \\
2 x+4 y-9 & =0
\end{aligned}
$
$\therefore$ Equation of the tangent is $2 \mathrm{x}-\mathrm{y}+1=0$ and equation of the normal is $2 \mathrm{x}+4 \mathrm{y}-9=0$
Question 3.
Find the equations of the tangents: to the parabola $y^2=6 x$, parallel to $3 x-2 y+5=0$

Solution:

$
y^2=6 x ; \quad 4 a=6 \quad \Rightarrow a=\frac{6}{4}=\frac{3}{2}
$
Equation of the tangent is $y=m x+\frac{a}{m}$
Here, the tangent is parallel to $3 x-2 y+5=0$.
$
\begin{gathered}
\therefore \quad \text { Slope of the tangent }=\text { slope of the line }=-\left(\frac{3}{-2}\right)=\frac{3}{2}=m \\
\text { and } a=\frac{3}{2}
\end{gathered}
$
So equation of the tangent is $y=\frac{3}{2} x+\frac{\frac{3}{2}}{\frac{3}{2}}$
$
y=\frac{3}{2} x+1 \Rightarrow 2 y=3 x+2
$
$\therefore$ Equation of the tangent is $3 x-2 y+2=0$
Question 4.
Find the equations of the tangents: to the parabola $4 x^2-y^2=64$ Which are parallel to $10 x-3 y+9=0$.

Solution:
Slope of $10 x-3 y+9=0$ is $-\left(\frac{10}{-3}\right)=\frac{10}{3}=m$
$\therefore \quad$ Slope of its parallel $=m=\frac{10}{3}$
The given hyperbola is $4 x^2-y^2=64$.
i.e., $\frac{4 x^2}{64}-\frac{y^2}{64}=1 \Rightarrow \frac{x^2}{16}-\frac{y^2}{64}=1$
Comparing this with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ we get, $a^2=16, b^2=64$
Here, $m=\frac{10}{3}$
So the equation of the tangent is $y=m x \pm \sqrt{a^2 m^2-b^2}$
$
\text { i.e., } \begin{aligned}
y & =\frac{10}{3} x \pm \sqrt{16\left(\frac{10}{3}\right)^2-64} \\
\text { i.e., } \quad y & =\frac{10 x}{3} \pm \sqrt{16\left(\frac{100}{9}\right)-64} \\
y & =\frac{10 x}{3} \pm \sqrt{\frac{1600-576}{9}}=\frac{10 x}{3} \pm \sqrt{\frac{1024}{3}}=\frac{10 x \pm 32}{3} \\
3 y & =10 x \pm 32
\end{aligned}
$
Question 5.
Find the equation of the tangent from point $(2,-3)$ to the parabola $y^2=4 x$.
Solution:
$
y^2=4 x
$
Equation of the tangent to the parabola will be of the form $\mathrm{y}=m x+\frac{1}{m}$ The tangents pass through $(2,-3)$
$
\begin{aligned}
\Rightarrow-3 & =2 m+\frac{1}{m} \\
-3 m & =2 m^2+1 \\
2 m^2+3 m+1 & =0
\end{aligned}
$

$
\begin{aligned}
& m=\frac{-3 \pm \sqrt{9-8}}{4}=\frac{-3 \pm 1}{4} \\
& m=\frac{-3+1}{4}, \frac{-3-1}{4} \Rightarrow m=\frac{-1}{2},-1
\end{aligned}
$
When $m=\frac{-1}{2}$, the equation of the tangent is
$
\begin{aligned}
y & =\frac{-1}{2} x+\frac{1}{\frac{-1}{2}} ; \text { i.e., } y=\frac{-x}{2}-2 \\
2 y & =-x-4 \\
x+2 y+4 & =0
\end{aligned}
$
When $m=-1$, equation of the tangent is
$
y=-1 x+\frac{1}{-1} \Rightarrow y=-x-1 \Rightarrow x+y+1=0
$
The two tangents drawn from $(2,-3)$ are $x+y+1=0, x+2 y+4=0$
Question 6.
Find the equation of the tangents from the point $(2,-3)$ to the parabola $2 x^2-3 y^2=6$

Solution:
$
2 x^2-3 y^2=6
$

$
(\div \text { by } 6) \Rightarrow \frac{2 x^2}{6}-\frac{3 y^2}{6}=1 ; \quad \text { i.e., } \frac{x^2}{3}-\frac{y^2}{2}=1 ; a^2=3 ; b^2=2
$
Equation of the tangent is of the form $y=m x \pm \sqrt{a^2 m^2-b^2}$
$
\text { i.e., } \quad y=m x \pm \sqrt{3 m^2-2}
$
The tangents are drawn from $(1,-1) \Rightarrow-1=m \pm \sqrt{3 m^2-2}$
$
\text { i.e., }-(1+m)=\pm \sqrt{3 m^2-2}
$
Squaring on both sides, $(1+m)^2=3 m^2-2$
$
\begin{aligned}
& m^2+2 m+1-3 m^2+2=0 \\
& -2 m^2+2 m+3=0 \\
& \text { i.e., } \quad 2 m^2-2 m-3=0 \\
& m=\frac{2 \pm \sqrt{4+24}}{4}=\frac{2 \pm 2 \sqrt{7}}{4}=\frac{1}{2} \pm \frac{\sqrt{7}}{2} \\
& m_1=\frac{1+\sqrt{7}}{2}, m_2=\frac{1-\sqrt{7}}{2} \\
&
\end{aligned}
$
When $m_1=\frac{1+\sqrt{7}}{2}$, equation of the the tangent is
$
\begin{aligned}
y & =\left(\frac{1+\sqrt{7}}{2}\right) x+\sqrt{3\left(\frac{1+\sqrt{7}}{2}\right)^2-2} \\
y & =\left(\frac{1+\sqrt{7}}{2}\right) x+\sqrt{\frac{3+6 \sqrt{7}+21-8}{2}} \\
2 y & =(1+\sqrt{7}) x+\sqrt{16+6 \sqrt{7}}
\end{aligned}
$
Similarly by substituting $m=\frac{1-\sqrt{7}}{2}$, equation of the the tangent is
$
2 y=(1-\sqrt{7}) x+\sqrt{16-6 \sqrt{7}}
$

Question 7.
Prove that the line $5 x+12 y=9$ touches the hyperbola $x^2-9 y^2=9$ and find the point of contact.

Solution:
The given line is $5 x+12 y=9 \Rightarrow y=\frac{-5}{12} x+\frac{9}{12}$
Comparing the above line with $y=m x+c$ we get,
$
m=\frac{-5}{12} \text { and } c=\frac{9}{12}=\frac{3}{4}
$
The given hyperbola is $x^2-9 y^2=9$
i.e., $\quad \frac{x^2}{9}-\frac{y^2}{1}=1 ;$ Here, $a^2=9 ; b^2=1$
Condition for the line $y=m x+c$ to be a tangent to the hyperbola
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } c^2=a^2 m^2-b^2 \text {. }
$
Here, $c^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}$
$
a^2 m^2-b^2=9\left(\frac{-5}{12}\right)^2-1=9\left(\frac{25}{144}\right)-1=\frac{25}{16}-1=\frac{9}{16}
$
(1) $=(2) \Rightarrow$ The given line is a tangent to the curve i.e., the given line touches the curve. To find the point of contact we have to solve the line and the curve.
The given line $5 x+12 y=9$

$\Rightarrow 12 y=9-5 x \Rightarrow y=\frac{9-5 x}{12}$
Substituting $y=\frac{9-5 x}{12}$ in $x^2-9 y^2=9$ we get,
$
\begin{aligned}
x^2-9\left[\frac{9-5 x}{12}\right]^2=9 & \\
x^2-\frac{9}{144}\left[81+25 x^2-90 x\right]=9 & \Rightarrow x^2-\frac{1}{16}\left[81+25 x^2-90 x\right]=9 \\
\frac{16 x^2-\left(81+25 x^2-90 x\right)}{16}=9 & \Rightarrow 16 x^2-81-25 x^2+90 x=144 \\
16 x^2-81-25 x^2-90 x-144 & =0 \Rightarrow-9 x^2+90 x-225=0 \\
(\div \text { by }-9) \Rightarrow x^2-10 x+25 & =0 \\
(x-5)^2 & =0 \Rightarrow x=5 \\
\text { when } x=5, y & =\frac{9-25}{12}=\frac{-16}{12}=\frac{-4}{3}
\end{aligned}
$
Point of contact is $\left(5, \frac{-4}{3}\right)$.

Question 8.
Show that the line $x-y+4=0$ is a tangent to the ellipse $x^2+3 y^2=12$. Find the co-ordinates of the point of contact.
Solution:
The given ellipse is $\mathrm{x}^2+3 \mathrm{y}^2=12$
Given line is $x-y+4=0 \Rightarrow y=x+4$
Here, $\mathrm{m}=1$ 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 $c^2=a^2 m^2+b^2$.
LHS: $c^2=4^2=16$
RHS: $\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}=12(1)+4=16$
LHS: RHS $\Rightarrow$ the given lines is a tangent to the ellipse.