Let the normal at the point P on the parabola y2 = 6x pass through the point (5, –8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is
-3
\(-\frac{9}{4}\)
\(-\frac{5}{2}\)
-2
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(-\frac{9}{4}\)
Let P(at2, 2at) where \(a=\frac{3}{2}\)
T :yt = x + at2 So point Q is
\((−at, at-\frac{a}{t})\)
N :y = –tx + 2at + at3 passes through (5, –8)
\(−8=−5t+3t+\frac{3}{2}t^3\)
⇒ 3t3 – 4t + 16 = 0
⇒ (t + 2)(3t2 – 6t + 8) = 0
⇒ t = –2
So ordinate of point Q is \(-\frac{9}{4}\)
Top Questions on Parabola
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.