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Tangent To a Parabola

The line 'l1'...

Question

The line 'l_1' passing through the point $(1, 1)$ and the 'l_2' passes through the point $(- 1, 1)$ . If the difference of the slope of lines is $2$ . Find the locus of the point of intersection of the $l_{1}$ and $l_{2}$ .

x^{2} = y

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y = 2 - x^{2}

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y^{2} = x

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x = 2 - y^{2}

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Solution

The correct option is A $x^{2} = y$
REF.Image.
Slope of $L_{1} = m_{1}$
slope of $L_{2} = m_{2}$
Let P(h,k) $\Rightarrow m_{1} = \frac{k - 1}{h - 1}$
$m_{2} = \frac{k - 1}{h + 1}$
Given
$m_{1} - m_{2} = 2$
$\Rightarrow \frac{k - 1}{h - 1} - \frac{k - 1}{h + 1} = 2$
$\Rightarrow (k - 1) [\frac{1}{h - 1} - \frac{1}{h + 1}] = 2$
$\Rightarrow (k - 1) [\frac{2}{h^{2} - 1}] = x$
$\Rightarrow h^{2} - 1 = k - 1 \Rightarrow h^{2} = k$
Locus of P(h,k) is
$x^{2} = y$

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