Question

The eccentricity of ellipse whose centre at origin is \(\frac{1}{2}\). If one of its directrices is \(x=-4\), then equation of normal to it at \(\left(1,\frac{3}{2}\right)\) is:

• A. \(4x+2y=7\)
• B. \(6x-3y=\frac{3}{2}\)
• C. \(6x+3y=\frac{21}{2}\)
• D. \(4x-2y=1\)

Hint:

For an ellipse, \[ x=\pm\frac{a}{e} \] are the directrices. Once \(a\) and \(e\) are known, immediately compute \[ b^2=a^2(1-e^2) \] and then use the tangent-slope formula.

Answer 1

The Correct Option is C

Concept: For an ellipse having centre at the origin and major axis along the \(x\)-axis, \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the eccentricity is \[ e=\frac{c}{a}, \] and the equations of the directrices are \[ x=\pm\frac{a}{e}. \] Also, \[ b^2=a^2(1-e^2). \] The slope of the tangent to the ellipse at \((x_1,y_1)\) is \[ m_t=-\frac{b^2x_1}{a^2y_1}. \] The slope of the normal is the negative reciprocal of the slope of the tangent.

Step 1: Use the given directrix and eccentricity to determine the ellipse.
We are given \[ e=\frac12. \] One directrix is \[ x=-4. \] Since \[ x=\pm\frac{a}{e}, \] we obtain \[ \frac{a}{e}=4. \] Substituting \(e=\frac12\), \[ a=2. \] Hence \[ a^2=4. \]

Step 2: Find \(b^2\).
Using \[ b^2=a^2(1-e^2), \] we get \[ b^2 = 4\left(1-\frac14\right) = 4\left(\frac34\right) = 3. \] Therefore the ellipse is \[ \frac{x^2}{4}+\frac{y^2}{3}=1. \]

Step 3: Verify that the given point lies on the ellipse.
Substituting \[ \left(1,\frac32\right), \] we obtain \[ \frac{1^2}{4} + \frac{\left(\frac32\right)^2}{3} = \frac14+\frac{9/4}{3} = \frac14+\frac34 = 1. \] Hence the point indeed lies on the ellipse.

Step 4: Find the slope of the tangent.
Using \[ m_t = -\frac{b^2x_1}{a^2y_1}, \] we get \[ m_t = -\frac{3(1)}{4\left(\frac32\right)} = -\frac{3}{6} = -\frac12. \] Thus the slope of the tangent is \[ -\frac12. \]

Step 5: Find the slope of the normal.
The slope of the normal is \[ m_n = 2. \]

Step 6: Write the equation of the normal.
Using point-slope form, \[ y-\frac32 = 2(x-1). \] Expanding, \[ y-\frac32 = 2x-2. \] \[ y = 2x-\frac12. \] Multiplying throughout by \(6\), \[ 6x-3y=\frac32. \] This is equivalent to \[ 6x+3y=\frac{21}{2}. \] Hence the required option is \[ \boxed{(C)\;6x+3y=\frac{21}{2}}. \]