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

Step 1: Identify the parameters of the parabola.
The equation is \((y-2)^2 = 3(x-1)\). This is a shifted parabola of the form \(Y^2 = 4aX\), where \(Y=y-2\), \(X=x-1\), and \(4a=3 \implies a=3/4\).
The vertex is at \((h,k) = (1,2)\). The focus is at \((h+a, k) = (1+3/4, 2) = (7/4, 2)\).
Step 2: Find the coordinates of the ends of the latus rectum.
The ends of the latus rectum are at a distance of \( \pm 2a \) from the focus, in the direction perpendicular to the axis.
The x-coordinate is the same as the focus: \(x=7/4\).
The y-coordinates are \(y = k \pm 2a = 2 \pm 2(3/4) = 2 \pm 3/2\).
The two ends are \((7/4, 2+3/2) = (7/4, 7/2)\) and \((7/4, 2-3/2) = (7/4, 1/2)\).
The condition \(q>3\) (which is \(3.5>3\)) means we choose the point \(L(p,q) = (7/4, 7/2)\).
Step 3: Find the equation of the tangent at this point.
The equation of the tangent to \(Y^2=4aX\) at \((X_1, Y_1)\) is \(YY_1 = 2a(X+X_1)\).
Here, \( (x_1, y_1) = (7/4, 7/2) \). So, \(X_1 = x_1-h = 7/4-1 = 3/4\) and \(Y_1 = y_1-k = 7/2-2 = 3/2\).
Substituting into the tangent equation: \( (y-2)(3/2) = 2(3/4)( (x-1) + 3/4 ) \).
\( (y-2)(3/2) = (3/2)(x - 1/4) \).
Cancel \(3/2\) from both sides: \(y-2 = x - 1/4\).
\( y = x + 2 - 1/4 = x + 7/4 \).
Rearranging the equation: \( 4y = 4x + 7 \implies 4x - 4y + 7 = 0 \).