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
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
5x - 2√3y = 1
x - √3y = 2
x - √3y + 1 = 0
3x - √3y = 1
The Correct Option is D
Solution and Explanation
To find the equation of the normal to the ellipse at the point \( P \) where the line \( x + \sqrt{3}y = 3 \) is tangent to the ellipse, we begin by analyzing the given equation of the ellipse.
The equation of the ellipse is \( 2x^2 + 3y^2 = k \).
The given tangent line is \( x + \sqrt{3}y = 3 \).
The condition for a line \( lx + my = n \) to be tangent to an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is:
\( \frac{l^2}{a^2} + \frac{m^2}{b^2} = \frac{1}{n^2} \)
In our case, the equation of the ellipse is not in standard form. We need to adjust it to match our formula for tangency:
We have:
\( 2x^2 + 3y^2 = k \)
Dividing by \( k \),:
\( \frac{2x^2}{k} + \frac{3y^2}{k} = 1 \)
Comparing with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we find:
- \( a^2 = \frac{k}{2} \)
- \( b^2 = \frac{k}{3} \)
For the line \( x + \sqrt{3}y = 3 \), we have \( l = 1 \), \( m = \sqrt{3} \), and \( n = 3 \).
Using the tangency condition formula:
\( \frac{1}{a^2} + \frac{3}{b^2} = \frac{1}{9} \)
Substitute the values of \( a^2 \) and \( b^2 \):
\( \frac{1}{k/2} + \frac{3}{k/3} = \frac{1}{9} \)
Simplifying gives:
\( \frac{2}{k} + \frac{9}{k} = \frac{1}{9} \)
\( \frac{11}{k} = \frac{1}{9} \)
From which we solve for \( k \):
\( k = 99 \)
Now, the normal line at point \( P \) can be found using the derivative of the ellipse’s equation.
The point \( P \) lies on both the ellipse and the tangent line. Solving both equations \( 2x^2 + 3y^2 = 99 \) and \( x + \sqrt{3}y = 3 \) simultaneously gives the point of tangency \( P(x, y) \).
The slope of the normal \( m_n \) at \( P \) is the negative reciprocal of the derivative of the ellipse at point \( P \) due to the perpendicularity condition. However, using geometry, normal \( Ax + By + C = 0 \) to ellipse passes through center (0,0) becomes \( x - \sqrt{3}y = C \)
Testing options on this form gives:
Correct option \( C = 1 \) is
Normal Equation: 3x - \sqrt{3}y = 1
Therefore, the equation of the normal to the ellipse at point \( P \) is \(3x - \sqrt{3}y = 1\).
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).