Question

If the line x – 2y = 12 is tangent to the ellipse \(\frac{{{\text{x}}^{2}}}{{{\text{a}}^{2}}}+\frac{{{\text{y}}^{2}}}{{{\text{b}}^{2}}}=1\text{ }\!\!~\!\!\text{ at }\!\!~\!\!\text{ the }\!\!~\!\!\text{ point }\!\!~\!\!\text{ }\left( 3,\frac{-9}{2} \right)\), then the length of the latus rectum of the ellipse is:
1. 9
2. 12√2
3. 5
4. 8√3
5. None of these

Answer 1

Correct Answer - Option 1 : 9

Concept:

Equation of tangent to the ellipse:\(\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) at the point (x₁, y₁) is given by: \(\frac{{x\; ⋅ {x_1}}}{{{a^2}}} + \frac{{y\; ⋅ {y_1}}}{{{b^2}}} = 1\)

The length of latus rectum of ellipse \(\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\), where a > b is given by: 2 ⋅ (b² / a)

Calculation:

Given: Equation of ellipse \(\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) and equation of tangent x – 2y = 12 at the point (3, -9/2)

As we know that equation of tangent to the ellipse:\(\;\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) at the point (x1, y1) is given by: \(\frac{{x\; ⋅ {x_1}}}{{{a^2}}} + \frac{{y\; ⋅ {y_1}}}{{{b^2}}} = 1\).

Here, x₁ = 3, y₁ = - 9/2.

The equation of tangent to \(\frac{{{\text{x}}^{2}}}{{{\text{a}}^{2}}}+\frac{{{\text{y}}^{2}}}{{{\text{b}}^{2}}}=1\text{ }\!\!~\!\!\text{ at}\left( 3,-\frac{9}{2} \right)\) is

\(\frac{{3{\rm{x}}}}{{{{\rm{a}}^2}}} - \frac{{9{\rm{y}}}}{{2{{\rm{b}}^2}}} = 1\)        ----(1)

Given that the equation of tangent is, x – 2y = 12.

The above equation can be re-written as: \(\frac{{x\;}}{{12}} - \frac{{y\;}}{6} = 1\)      ----(2)

Now, by comparing equation (1) and (2), we get: \(\frac{3}{{{{\rm{a}}^2}}} = \frac{1}{{12}}\) and \(\frac{{ - 9}}{{ 4{{\rm{b}}^2}}} = \frac{-1}{{12}}\)

\(\Rightarrow \frac{3}{{{{\rm{a}}^2}}} = \frac{1}{{12}}\)

⇒ a² = 3 × 12

⇒ a² = 36

∴ a = 6

Similarly, 

\(\Rightarrow \frac{{ - 9}}{{ 4{{\rm{b}}^2}}} = \frac{-1}{{12}}\)

\(\Rightarrow {{\rm{b}}^2} = \frac{{9 \times 12}}{4}\)

⇒ b² = 9 × 3 = 27

\(\therefore {\rm{b}} = 3\sqrt 3\)

∵ a > b and we know that the length of latus rectum of ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\), where a > b is given by: 2 ⋅ (b2 / a)

\(\Rightarrow \frac{{2{{\rm{b}}^2}}}{{\rm{a}}} = \frac{{2 \times 27}}{6} = 9\)