Let T1 and T2 be two distinct common tangents to the ellipse \(E: \frac{x^2}{6} + \frac{y^2}{3} = 1\) and the parabola \(P: y^2 = 12x\). Suppose that the tangent T1 touches P and E at the points A1 and A2, respectively and the tangent T2 touches P and E at the points A4 and A3, respectively. Then which of the following statements is(are) true?
The area of the quadrilateral A1A2A3A4 is 35 square units
The area of the quadrilateral A1A2A3A4 is 36 square units
The tangents T1 and T2 meet the x -axis at the point (–3,0)
The tangents T1 and T2 meet the x -axis at the point (–6,0)
The Correct Option is A, C
Solution and Explanation
Given :
\(E:\frac{x^2}{6}+\frac{y^2}{3}=1\), Tangent : \(y=m_1x±\sqrt{6m_1^2+3}\)
P : y2 = 12x, Tangent : \(y=m_2x+\frac{3}{m_2}\)
Now, for common tangent :
\(m=m_1+m_2,±\sqrt{6m_1^2+3}=\frac{3}{m_2}\)
⇒ m = ±1
The equations of the common tangents are y = x + 3 and y = -x - 3. The points of contact for the parabola are:
\(\left(\frac{a}{m^2},\frac{2a}{3}\right)\)
A1 ≡ (3, 6), A4(3 - 6)
Now, Let suppose A2(x1, y1)
⇒ tangent to E : \(\frac{xx_1}{6}+\frac{yy_1}{3}=1\)
A3 is mirror image of A2 in x-axis ⇒ A3(–2, –1)
The intersection point of T1 = 0 and T2 = 0 is at (-3, 0).
Area of quadrilateral A1A2A3A4 :
\(\frac{1}{2}(12+2)\times5=35\) square units.
So, the correct options are (A) and (C).
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Conic Sections
When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
Let ‘β’ is the angle made by the plane with the vertical axis of the cone
- When β = 90°, we say the section is a circle
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