The intercepts of the focal chord (which is a | vedclass

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(D) Let the focal chord be $PQ$ passing through the focus $S(a, 0)$. Let the coordinates of $P$ be $(at_{1}^{2}, 2at_{1})$ and $Q$ be $(at_{2}^{2}, 2a

Vedclass · Accepted Answer

$\frac{ab}{b-a}$

Vedclass · Answer

(D) Let the focal chord be $PQ$ passing through the focus $S(a, 0)$. Let the coordinates of $P$ be $(at_{1}^{2}, 2at_{1})$ and $Q$ be $(at_{2}^{2}, 2at_{2})$.Since it is a focal chord,$t_{1}t_{2} = -1$.The focal distance of any point $P(x, y)$ on the parabola is $SP = a + x = a + at_{1}^{2} = a(1 + t_{1}^{2})$.Given the intercepts of the focal chord are $b$ and $k$,we have $b = a(1 + t_{1}^{2})$ and $k = a(1 + t_{2}^{2})$.Since $t_{2} = -1/t_{1}$,we have $k = a(1 + (-1/t_{1})^{2}) = a(1 + 1/t_{1}^{2}) = a(\frac{t_{1}^{2} + 1}{t_{1}^{2}})$.From $b = a(1 + t_{1}^{2})$,we get $t_{1}^{2} = \frac{b}{a} - 1 = \frac{b-a}{a}$.Substituting this into the expression for $k$:$k = a(1 + \frac{1}{(b-a)/a}) = a(1 + \frac{a}{b-a}) = a(\frac{b-a+a}{b-a}) = a(\frac{b}{b-a}) = \frac{ab}{b-a}$.

List-$I$	List-$II$
$A$. Equation of the tangent drawn at $(2, \sqrt{8})$ on the curve $y^2 = 4x$ is	$(i) -36$
$B$. Equation of the normal to the curve $y^2 = 16x$,that makes an angle of $45^{\circ}$ with its axis is	$(ii) 4$
$C$. The chord joining the points $(x_1, y_1)$ and $(x_2, y_2)$ on the curve $y^2 = 12x$ is a focal chord if $y_1 y_2 =$	$(iii) 8$
$D$. $A$ value of $k$ for which $x - 3 = 0$ is the directrix of the curve $y^2 - kx + 16 = 0$ is	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

The intercepts of the focal chord (which is a part of the latus rectum) to the parabola $y^{2}=4ax$ are $b$ and $k$. Then $k$ is equal to:

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