Parabola Tangent: GP Relation for Fixed Point Chords

[DaalChawal]

Tangent is drawn at any point ( $x_1$ , $y_1$ ) other than vertex on the parabola $y^2$ = 4ax . If tangents are drawn from any point on this tangent to the circle $x^2$ + $y^2$ = $a^2$ such that all chords of contact pass through a fixed point ( $x_2$ , $y_2$ ) then
(A) $x_1$ , a , $x_2$ are in G.P.
(B) $y_{1} \over 2$ ,a, $y_2$ are in G.P.
(C) -4 , $y_{1} \over y_{2}$ , $x_{1} \over x_{2}$ are in G.P.
(D) $x_1$ $x_2$ + $y_1$ $y_2$ = $a^2$

 

[skeeter]

Tangent is drawn at any point $( x1,y_1 )$ other than vertex on the parabola $y^2 = 4ax$.

let $(x_1,y_1) = (a, 2a)$

slope of the tangent line at $(a,2a)$ is $m = 1$

tangent line equation is $y - 2a = x-a \implies y = x+a$

$y = x+a$ intersects the circle $x^2+y^2 = a^2$ when $x^2 + (x+a)^2 = a^2 \implies 2x(x+a) = 0 \implies x_2 = 0 \text{ or } x_2 = -a$

$x_2 = 0 \implies y_2 = a$

(A) $\{x_1, a, x_2\} = \{a, a, 0\}$

(B) $\{y_1/2, a, y_2\} = \{a, a, a\}$

(C) $\{-4, y_1/y_2, x_1/x_2 \} = \{-4, 2, \emptyset \}$

(D) $x_1x_2 + y_1y_2 = a \cdot 0 + 2a \cdot a = 2a^2$now, check the four choices for $x_2=-a$

 

[DaalChawal]

If tangents are drawn from any point on this tangent to the circle x2x2x^2 + y2y2y^2 = a2a2a^2

y=x+ay=x+ay = x+a intersects the circle x2+y2=a2x2+y2=a2x^2+y^2 = a^2

Here the question says that from any point on the tangent to parabola, a tangent to the circle is drawn and that tangent is the chord of contact of parabola which passes through ( $x_2$ , $y_2$ ). And you have taken that tangent of parabola as chord of contact of circle also question says ( $x_2$ , $y_2$ ) is a fixed point that does not mean it lies on circle. Do correct me if I'm wrong.

 

[skeeter]

So, are there two different tangent lines, one tangent to the parabola and the other tangent to the circle?

Has this problem been translated to English?

Maybe you can post your interpretation with a sketch?

 

[DaalChawal]

Yes, there are two different tangent lines.

Has this problem been translated to English?

No, you can look it's the question no. 255

This is what I'm saying