Common Tangents to 4(x²+y²)=9 and y²=4x Meet at Q — Find l/e² of an Ellipse

JEE Main 2022 — 26 June, Morning Shift. Previous Year Question.

Problem

Let the common tangents to the curves $4(x^2+y^2)=9$ and $y^2=4x$ intersect at the point $Q$. Let an ellipse, centred at the origin $O$, have lengths of semi-minor and semi-major axes equal to $OQ$ and $6$, respectively. If $e$ and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\dfrac{l}{e^2}$ is equal to ____.

Key insight. Instead of working with the parabola’s tangent equation and testing it against the circle, write the tangent to the parabola in slope form (which has just one parameter, the slope $m$), then apply the circle’s “perpendicular distance from centre equals radius” condition. That produces an equation purely in $m$, giving both tangent lines directly.

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Approach

Write the tangent to $y^2=4x$ in slope form as $y=mx+\frac{1}{m}$, then impose the condition that this line is also tangent to the circle $x^2+y^2=\frac{9}{4}$ (its perpendicular distance from the origin equals the radius). Solve for $m$, find both tangent lines, locate their intersection point $Q$, and use $OQ$ as the ellipse’s semi-minor axis to compute $l/e^2$.

Solution

Step 1 — Write the tangent to the parabola

For $y^2=4x$ (so $4a=4$, $a=1$), a tangent with slope $m$ has equation:

$$y = mx+\frac{1}{m}$$

Step 2 — Rewrite the circle in standard form

$$4(x^2+y^2)=9 \implies x^2+y^2=\frac{9}{4}$$

This is a circle centred at the origin with radius $\dfrac{3}{2}$.

Step 3 — Apply the tangent condition to the circle

The perpendicular distance from the origin to $y-mx-\frac{1}{m}=0$ must equal the radius:

$$\frac{\left|-\frac{1}{m}\right|}{\sqrt{1+m^2}} = \frac{3}{2}$$

Squaring:

$$\frac{1}{m^2(1+m^2)} = \frac{9}{4} \implies 4 = 9m^2(1+m^2) \implies 9m^4+9m^2-4=0$$

Step 4 — Solve for m²

Let $u=m^2$:

$$9u^2+9u-4=0 \implies u = \frac{-9\pm\sqrt{81+144}}{18} = \frac{-9\pm15}{18}$$

$$u = \frac{1}{3} \text{ or } u=-\frac{4}{3} \text{ (rejected, negative)}$$

So $m^2=\dfrac{1}{3}$, giving $m=\pm\dfrac{1}{\sqrt3}$.

Step 5 — Write both tangent lines

$$y = \frac{x}{\sqrt3}+\sqrt3 \qquad \text{and} \qquad y = -\frac{x}{\sqrt3}-\sqrt3$$

Step 6 — Find their intersection point Q

Setting the two equal:

$$\frac{x}{\sqrt3}+\sqrt3 = -\frac{x}{\sqrt3}-\sqrt3 \implies \frac{2x}{\sqrt3}=-2\sqrt3 \implies x=-3$$

Substituting back: $y = -\dfrac{3}{\sqrt3}+\sqrt3 = -\sqrt3+\sqrt3=0$. So $Q=(-3,0)$.

Step 7 — Set up the ellipse

$OQ=3$, so the semi-minor axis $b=3$. The semi-major axis is given as $a=6$.

$$c^2 = a^2-b^2 = 36-9=27$$

Step 8 — Compute l and e²

$$l = \frac{2b^2}{a} = \frac{2(9)}{6}=3, \qquad e^2 = \frac{c^2}{a^2} = \frac{27}{36}=\frac{3}{4}$$

Step 9 — Compute l/e²

$$\frac{l}{e^2} = \frac{3}{\frac{3}{4}} = 4$$

Answer

Common mistakes

• Using the tangent condition for the parabola on the circle’s equation, or vice versa. The slope-form tangent $y=mx+\frac{1}{m}$ is specific to the parabola; it must then be tested for tangency against the circle separately using the perpendicular-distance formula.
• Forgetting to reject the negative value of $m^2$. The quadratic in $u=m^2$ has one negative root, which is impossible for a real slope — only the positive root is valid.

Practise next

• Let the common tangents to $x^2+y^2=4$ and $y^2=8x$ intersect at point $R$. An ellipse centred at the origin has semi-minor axis $OR$ and semi-major axis $8$. Find $l/e^2$, using the same slope-form tangent method.