Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of π/4 with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is
- 8 only
- 2 only
\(\frac{1}{4}\) only
- any a > 0
The Correct Option is D
Solution and Explanation
The correct answer is (D) : any a > 0
\(P:y²=4ax, a > 0 \) \(S(a,0)\)
Equation of tangent on parabola
\(y = mx + \frac{a}{m}\)
y = 3x + 5
\(tan \frac{π}{4} = |\frac{m-3}{1+3m} | ⇒ m-3 = ± ( 1+3m )\)
m-3 = 1+3m
m=-2
m-3 = -1-3m
\(m=\frac{1}{2}\)
Equation of one tangent :\( y = -2x - \frac{a}{2}\)
Equation of other tangent : \(y = \frac{x}{2} + 2a\)
Point of contact are
\(( \frac{a}{(-2)²} , \frac{-2a}{(-2)} ) and ( \frac{a}{(\frac{1}{2})²} , \frac{-2a}{\frac{1}{2}} )\)
\(A ( \frac{a}{4},a)\) and \(B (4a,-4a)\)
Now or \((ΔABS)\) = 0 [ S is the focus ]
\(\frac{1}{2} \begin{vmatrix} \frac{a}{4} & a & 1 \\ 4a & -4a & 1 \\ a& 0 & 1 \end{vmatrix} = 0\)
\(⇒ \frac{a}{4} (-4a-0) -a(4a-a) + 1(0-(-4a²)) = 0\)
\(= -a² -3a² + 4a² = 0\)
Always true
Top Questions on Parabola
- If the line \( 3x - 2y + 12 = 0 \) intersects the parabola \( 4y = 3x^2 \) at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to:
- Let A and B be the two points of intersection of the line \( y + 5 = 0 \) and the mirror image of the parabola \( y^2 = 4x \) with respect to the line \( x + y + 4 = 0 \). If \( d \) denotes the distance between A and B, and \( a \) denotes the area of \( \Delta SAB \), where \( S \) is the focus of the parabola \( y^2 = 4x \), then the value of \( (a + d) \) is:
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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.