SATHEE: Parabola 1 Question 13

Parabola 1 Question 13

13. Let the curve $C$ be the mirror image of the parabola $y^{2} = 4 x$ with respect to the line $x + y + 4 = 0$ . If $A$ and $B$ are the points of intersection of $C$ with the line $y = - 5$ , then the distance between $A$ and $B$ is

(2015 Adv.)

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Answer:

Correct Answer: 13. (4)

Solution:

Let $P (t^{2}, 2 t)$ be a point on the curve $y^{2} = 4 x$ , whose image is $Q (x, y)$ on $x + y + 4 = 0$ , then

$\begin{aligned} \frac{x - t^{2}}{1} & = \frac{y - 2 t}{1} = \frac{- 2 (t^{2} + 2 t + 4)}{1^{2} + 1^{2}} \\ \Rightarrow & x & = - 2 t - 4 \\ and & y & = - t^{2} - 4 \end{aligned}$

Now, the straight line $y = - 5$ meets the mirror image.

$\begin{array}{lc} ∴ & - t^{2} - 4 = - 5 \\ \Rightarrow & t^{2} = 1 \\ \Rightarrow & t = \pm 1 \end{array}$

Thus, points of intersection of $A$ and $B$ are $(- 6, - 5)$ and $(- 2, - 5)$ .

$∴$ Distance, $A B = \sqrt{(- 2 + 6)^{2} + (- 5 + 5)^{2}} = 4$

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Parabola 1 Question 13

13. Let the curve C be the mirror image of the parabola y2=4x with respect to the line x+y+4=0. If A and B are the points of intersection of C with the line y=−5, then the distance between A and B is

Answer:

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13. Let the curve $C$ be the mirror image of the parabola $y^{2} = 4 x$ with respect to the line $x + y + 4 = 0$ . If $A$ and $B$ are the points of intersection of $C$ with the line $y = - 5$ , then the distance between $A$ and $B$ is