Question

If L(p,q), q>3 is one end of the latus rectum of the parabola \((y-2)^2 = 3(x-1)\) then the equation of the tangent at L to this parabola is

• A. \( 2x+y-7=0 \)
• B. \( 4x-4y+7=0 \)
• C. \( 2x-y-3=0 \)
• D. \( 2x-3y+7=0 \)

Hint:

The tangent at one end of the latus rectum of a parabola intersects the directrix on the axis of the parabola. Also, the tangent at the end of the latus rectum is inclined at 45 degrees to the axis.

Answer 1

The Correct Option is B

To solve the given problem, we first need to understand the parabola's properties and equations involved.

The given equation of the parabola is \((y-2)^2 = 3(x-1)\).

1. This is in the form \((y-k)^2 = 4a(x-h)\), which represents a parabola that opens horizontally.
2. From the equation \((y-2)^2 = 3(x-1)\), we can deduce:
   • The vertex of the parabola is at \((h,k) = (1,2)\).
   • The value of \(4a = 3\) implies \(a = \frac{3}{4}\).
3. The latus rectum of a parabola is a line segment parallel to the directrix and through the focus, with endpoints on the parabola. Its length is \(|4a| = 3\).
   • For a parabola with equation \((y-k)^2 = 4a(x-h)\), the length of the latus rectum is \(|4a|\), and it passes through the focus.
   • The focus of the parabola is at \((h+a, k) = \left(1 + \frac{3}{4}, 2\right) = \left(\frac{7}{4}, 2\right)\).
   • Thus, the endpoints of the latus rectum are \(\left(\frac{7}{4}, 2+ \frac{3}{2}\right)\) and \(\left(\frac{7}{4}, 2 - \frac{3}{2}\right)\) which evaluates to \(\left(\frac{7}{4}, \frac{7}{2}\right)\) and \(\left(\frac{7}{4}, \frac{1}{2}\right)\), respectively.
4. We are given one end of the latus rectum, \(L(p,q)\), where \(q > 3\).
   • This means \(L\left(\frac{7}{4}, \frac{7}{2}\right)\) is the point as \(\frac{7}{2} = 3.5 > 3\).

Now, we find the equation of the tangent to the parabola at this point \(L\left(\frac{7}{4}, \frac{7}{2}\right)\).

1. The equation of the tangent to the parabola \((y-k)^2 = 4a(x-h)\) at point \((x_1, y_1)\) is given by:
   • \(y - y_1 = \frac{2a}{y_1-k} (x - x_1)\)
2. Substituting the given values and the point into the formula:
   • \(y - \frac{7}{2} = \frac{\frac{3}{2}}{\frac{7}{2} - 2} (x - \frac{7}{4})\)
   • Which simplifies to: \(y - \frac{7}{2} = \frac{3}{2} (x - \frac{7}{4})\)
   • Further simplification leads to: \(2(y - \frac{7}{2}) = 3(x - \frac{7}{4})\)
   • Finally, obtain the tangent equation: \(4x - 4y + 7 = 0\)

Thus, the equation of the tangent at point \(L(p,q)\) is \(4x - 4y + 7 = 0\), which matches the correct answer choice.

Conclusion: The correct answer is \( 4x-4y+7=0 \).