Parabola Ques 38
- The tangents to the curve $y=(x-2)^{2}-1$ at its points of intersection with the line $x-y=3$, intersect at the point
(a) $\frac{5}{2}, 1$
(b) $-\frac{5}{2},-1$
(c) $\frac{5}{2},-1$
(d) $-\frac{5}{2}, 1$
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Answer:
Correct Answer: 38.(c)
Solution:
- Given equation of parabola is
$$ \begin{aligned} & y=(x-2)^{2}-1 \\ & \Rightarrow \quad y=x^{2}-4 x+3 \end{aligned} $$
Now, let $\left(x _1, y _1\right)$ be the point of intersection of tangents of parabola (i) and line $x-y=3$, then
Equation of chord of contact of point $\left(x _1, y _1\right)$ w.r.t. parabola (i) is
$T=0$
$\Rightarrow \frac{1}{2}\left(y+y _1\right)=x x _1-2\left(x+x _1\right)+3$
$\Rightarrow y+y _1=2 x\left(x _1-2\right)-4 x _1+6$
$\Rightarrow 2 x\left(x _1-2\right)-y=4 x _1+y _1-6$, this equation represent the line $x-y=3$ only, so on comparing, we get
$$ \begin{aligned} & \frac{2\left(x _1-2\right)}{1}=\frac{-1}{-1}=\frac{4 x _1+y _1-6}{3} \\ \Rightarrow & x _1=\frac{5}{2} \text { and } y _1=-1 \end{aligned} $$
So, the required point is $\frac{5}{2},-1$.
BETA
