Question

If x+√3y = 3 is the tangent to the ellipse 2x² + 3y² = k at a point P then the equation of the normal to this ellipse at P is

• A. 5x - 2√3y = 1
• B. x - √3y = 2
• C. x - √3y + 1 = 0
• D. 3x - √3y = 1

Answer 1

The Correct Option is D

To find the equation of the normal to the ellipse \(2x^2 + 3y^2 = k\) at the point \(P\) where the line \(x + \sqrt{3}y = 3\) is tangent, we proceed as follows:

1. Substituting the Line into the Ellipse:
The line is \(x + \sqrt{3}y = 3\), so solve for \(x\):

\( x = 3 - \sqrt{3}y \)
Substitute into the ellipse equation \(2x^2 + 3y^2 = k\):

\( 2(3 - \sqrt{3}y)^2 + 3y^2 = k \)
Expand:

\( (3 - \sqrt{3}y)^2 = 9 - 6\sqrt{3}y + 3y^2 \)
\( 2(9 - 6\sqrt{3}y + 3y^2) + 3y^2 = 18 - 12\sqrt{3}y + 6y^2 + 3y^2 = 9y^2 - 12\sqrt{3}y + 18 \)
So:

\( 9y^2 - 12\sqrt{3}y + 18 = k \)
\( 9y^2 - 12\sqrt{3}y + (18 - k) = 0 \)

2. Tangency Condition:
Since the line is tangent to the ellipse, there is exactly one point of intersection, so the quadratic in \(y\) has one solution. The discriminant must be zero:

For \(9y^2 - 12\sqrt{3}y + (18 - k) = 0\), the discriminant is:

\( \Delta = (-12\sqrt{3})^2 - 4 \cdot 9 \cdot (18 - k) = 144 \cdot 3 - 36(18 - k) = 432 - 648 + 36k \)
Set \(\Delta = 0\):

\( 432 - 648 + 36k = 0 \)
\( -216 + 36k = 0 \)
\( 36k = 216 \implies k = 6 \)

3. Finding the Point of Tangency:
With \(k = 6\), the quadratic becomes:

\( 9y^2 - 12\sqrt{3}y + (18 - 6) = 9y^2 - 12\sqrt{3}y + 12 = 0 \)
Divide by 3:

\( 3y^2 - 4\sqrt{3}y + 4 = 0 \)
Rewrite:

\( (\sqrt{3}y - 2)^2 = 3y^2 - 4\sqrt{3}y + 4 = 0 \)
\( \sqrt{3}y - 2 = 0 \implies y = \frac{2}{\sqrt{3}} \)
Find \(x\):

\( x = 3 - \sqrt{3} \cdot \frac{2}{\sqrt{3}} = 3 - 2 = 1 \)
So, the point \(P\) is \(\left(1, \frac{2}{\sqrt{3}}\right)\).

4. Slope of the Tangent:
Differentiate the ellipse \(2x^2 + 3y^2 = 6\) implicitly:

\( 4x + 6y \frac{dy}{dx} = 0 \)
\( \frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y} \)
At \(P\left(1, \frac{2}{\sqrt{3}}\right)\):

\( \frac{dy}{dx} = -\frac{2 \cdot 1}{3 \cdot \frac{2}{\sqrt{3}}} = -\frac{2}{\frac{6}{\sqrt{3}}} = -\frac{2 \sqrt{3}}{6} = -\frac{1}{\sqrt{3}} \)

5. Slope of the Normal:
The slope of the normal is the negative reciprocal of the tangent’s slope:

\( m_{\text{normal}} = -\frac{1}{-\frac{1}{\sqrt{3}}} = \sqrt{3} \)

6. Equation of the Normal:
Using point-slope form at \(P\left(1, \frac{2}{\sqrt{3}}\right)\):

\( y - \frac{2}{\sqrt{3}} = \sqrt{3} (x - 1) \)
Multiply through by \(\sqrt{3}\):

\( \sqrt{3} y - 2 = 3(x - 1) \)
\( \sqrt{3} y - 2 = 3x - 3 \)
\( 3x - \sqrt{3} y - 3 + 2 = 0 \)
\( 3x - \sqrt{3} y - 1 = 0 \)
Thus:

\( 3x - \sqrt{3} y = 1 \)

Final Answer:
The equation of the normal to the ellipse at point \(P\) is \(3x - \sqrt{3} y = 1\).