JEE Conics Questions

We need to find the area of triangle PQR. To determine point P on the parabola $$y^2 = 3x$$, note that the tangent at point $$(at^2, 2at)$$ on $$y^2 = 4ax$$ has slope $$\frac{1}{t}$$. Here $$4a = 3$$, so $$a = \frac{3}{4}$$. The line $$x + 2y = 1$$ has slope $$-\frac{1}{2}$$. Thus $$\frac{1}{t} = -\frac{1}{2}$$, giving $$t = -2$$. It follows that $$P = \left(\frac{3}{4} \times 4, 2 \times \frac{3}{4} \times (-2)\right) = (3, -3)$$.

Next, we find points Q and R on the ellipse $$\frac{x^2}{4} + y^2 = 1$$. Lines perpendicular to $$x - y = 2$$ (slope 1) have slope $$-1$$. The general tangent to the ellipse is given by $$y = mx \pm \sqrt{a^2m^2 + b^2}$$ where $$a^2 = 4, b^2 = 1$$. Hence the tangents with slope -1 are $$y = -x \pm \sqrt{4 + 1} = -x \pm \sqrt{5}$$.

For the tangent $$y = -x + \sqrt{5}$$, substituting into the ellipse equation yields

$$\frac{x^2}{4} + (-x + \sqrt{5})^2 = 1$$

$$\frac{x^2}{4} + x^2 - 2\sqrt{5}x + 5 = 1$$

$$\frac{5x^2}{4} - 2\sqrt{5}x + 4 = 0$$

$$5x^2 - 8\sqrt{5}x + 16 = 0$$

$$x = \frac{8\sqrt{5} \pm \sqrt{320 - 320}}{10} = \frac{8\sqrt{5}}{10} = \frac{4\sqrt{5}}{5}$$.

Then

$$y = -\frac{4\sqrt{5}}{5} + \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$$, so $$Q = \left(\frac{4\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) = \left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$. Similarly, for the tangent $$y = -x - \sqrt{5}$$ one finds $$R = \left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$.

Finally, the area of triangle PQR with vertices $$P(3,-3)$$, $$Q\left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$, and $$R\left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$ is given by

Area $$= \frac{1}{2}|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$$

$$= \frac{1}{2}\left|3 \cdot \frac{2}{\sqrt{5}} + \frac{4}{\sqrt{5}}\left(-\frac{1}{\sqrt{5}} - (-3)\right) + \left(-\frac{4}{\sqrt{5}}\right)\left(-3 - \frac{1}{\sqrt{5}}\right)\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4}{\sqrt{5}} \cdot \frac{-1 + 3\sqrt{5}}{\sqrt{5}} - \frac{4}{\sqrt{5}} \cdot \frac{-3\sqrt{5} - 1}{\sqrt{5}}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4(-1+3\sqrt{5})}{5} + \frac{4(3\sqrt{5}+1)}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{-4+12\sqrt{5}+12\sqrt{5}+4}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24\sqrt{5}}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24}{\sqrt{5}}\right| = \frac{1}{2} \cdot \frac{30}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$.

Therefore, the area is $$\boxed{3\sqrt{5}}$$, which corresponds to Option D.

The curve $$4x^2 + 9y^2 = 36$$ can be written as $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, which is an ellipse with $$a^2 = 9$$ and $$b^2 = 4$$. The curve $$(2x)^2 + (2y)^2 = 31$$ simplifies to $$x^2 + y^2 = \frac{31}{4}$$, which is a circle with center at the origin and radius $$\frac{\sqrt{31}}{2}$$.

For the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, a tangent line with slope $$m$$ has the equation $$y = mx \pm \sqrt{a^2 m^2 + b^2} = mx \pm \sqrt{9m^2 + 4}$$. The perpendicular distance from the origin to this tangent is $$\frac{\sqrt{9m^2 + 4}}{\sqrt{1 + m^2}}$$.

For the circle $$x^2 + y^2 = \frac{31}{4}$$, a tangent line with slope $$m$$ has the equation $$y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1 + m^2}$$. The perpendicular distance from the origin to this tangent is $$\frac{\sqrt{31}}{2}$$.

For a common tangent, these distances must be equal (since both tangent lines pass at the same distance from the common center, the origin): $$\frac{\sqrt{9m^2 + 4}}{\sqrt{1 + m^2}} = \frac{\sqrt{31}}{2}$$.

Squaring both sides: $$\frac{9m^2 + 4}{1 + m^2} = \frac{31}{4}$$.

Cross-multiplying: $$4(9m^2 + 4) = 31(1 + m^2)$$, so $$36m^2 + 16 = 31 + 31m^2$$.

Simplifying: $$5m^2 = 15$$, hence $$m^2 = 3$$.