Parabola Ques 49
- Three normals are drawn from the point $(c, 0)$ to the curve $y^{2}=x$. Show that $c$ must be greater than $\frac{1}{2}$. One normal is always the $X$-axis. Find $c$ for which the other two normals are perpendicular to each other.
(1991, 4M)
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Answer:
Correct Answer: 49.$\frac{3}{4}$
Solution:
Formula:
- Let $A\left(t _1^{2}, 2 t _1\right)$ and $B\left(t _2^{2}, 2 t _2\right)$ be coordinates of the end points of a chord of the parabola $y^{2}=4 x$ having slope 2 .
Now, slope of $A B$ is
$$ m=\frac{2 t _2-2 t _1}{t _2^{2}-t _1^{2}}=\frac{2\left(t _2-t _1\right)}{\left(t _2-t _1\right)\left(t _2+t _1\right)}=\frac{2}{t _2+t _1} $$
But
$$ \begin{gathered} m=2 \\ 2=\frac{2}{t_2+t_1} \\ t₁ + t₂ = 1 \end{gathered} $$
[given]
Let $P(h, k)$ be a point on $A B$ such that, it divides $A B$ internally in the ratio $1: 2$.
Then, $\quad h=\frac{2 t_1^{2}+t_2^{2}}{2+1}$ and $k=\frac{2\left(2 t_1\right)+2 t_2}{2+1}$
$$ \begin{aligned} & \Rightarrow \quad 3 h=2 t_1^{2}+t_2^{2} \\ & \text { and } \quad 3 k=4 t _1+2 t _2 \end{aligned} $$
On substituting value of $t _1$ from Eq. (i) in Eq. (iii)
$$ \begin{array}{rlrl} & & 3 k & =4\left(1-t _2\right)+2 t _2 \\ \Rightarrow & 3 k & =4-2 t^2 \\ \Rightarrow & & t _2=2-\frac{3 k}{2} \end{array} $$
On substituting $t _1=1-t _2$ in Eq. (ii), we get
$$ \begin{aligned} & 3 h=2\left(1-t_2\right)^{2}+t_2^{2} \\ & =2\left(1-2 t_2+t_2^{2}\right)+t_2^{2} \\ & =3 t _2^{2}-4 t _2+2=3 \quad t _2^{2}-\frac{4}{3} t _2+\frac{2}{3} \\ & =3 \quad t_2-\frac{2}{3}^{2}+\frac{2}{3}-\frac{4}{9}=3 \quad t_2-\frac{2}{3}^{2}+\frac{2}{3} \\ & \Rightarrow \quad 3 h-\frac{2}{3}=3 \quad t_2-\left(\frac{2}{3}\right)^2 \ & \Rightarrow \quad 3 h-\frac{2}{9}=32-\frac{3 k}{2}-\left(\frac{2}{3}\right)^{2} \quad \text { [from Eq. (iv)] } \\ & \Rightarrow \quad 3 h-\frac{2}{9}=3 \cdot \frac{4}{3}-\left(\frac{3 k}{2}\right)^{2} \\ & \Rightarrow \quad h-\frac{2}{9}=\frac{9}{4} \quad \left(k-\frac{8}{9}\right)^{2} \\ & \Rightarrow \quad k-\left(\frac{8}{9}\right)^{2}=\frac{4}{9} \quad h-\frac{2}{9} \end{aligned} $$
434 Parabola
On generalizing, we get the required locus
$$ y-\frac{8}{9}^{2}=\frac{4}{9} \quad x-\frac{2}{9} $$
This represents a parabola with vertex at $\frac{2}{9}, \frac{8}{9}$.
BETA
