Question

The end of a latus rectum of the ellipse \(3x^2 + 4y^2 = 12\) is lying in the third quadrant. If the normal drawn at \(L_1\) to this ellipse intersects the ellipse again at the point \(P(a,b)\), then find the value of \(a\):

• A. \(\frac{63}{38}\)
• B. \(\frac{11}{19}\)
• C. \(\frac{-11}{19}\)
• D. \(\frac{-63}{38}\)

Hint:

For ellipses, always rewrite the equation in standard form.
- The normal to the ellipse at a point can be found using the equation for the tangent and its slope.

Answer 1

The Correct Option is B

To solve this problem, we first need to understand the given equation of the ellipse \(3x^2 + 4y^2 = 12\). We rearrange it to its standard form: \[ \frac{x^2}{4} + \frac{y^2}{3} = 1. \] From this, we can identify the semi-major axis \(a = 2\) and semi-minor axis \(b = \sqrt{3}\). For an ellipse, the equation for the latus rectum is given by \(y = k\), where \(k\) is the distance from the center to the focus. Using the properties of the latus rectum and normal to the ellipse at a given point, we can derive the value of \(a\). Solving the equations and applying the conditions, we find the value of \(a = \frac{11}{19}\). Thus, the correct answer is \(\boxed{\frac{11}{19}}\).