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Question:
If the tangents drawn at the points \(P\) and \(Q\) on the parabola \(y^2=2 x-3\) intersect at the point \(R (0,1)\), then the orthocentre of the triangle \(PQR\) is :
If the tangents drawn at the points \(P\) and \(Q\) on the parabola \(y^2=2 x-3\) intersect at the point \(R (0,1)\), then the orthocentre of the triangle \(PQR\) is :
Updated On: Sep 12, 2024
- \((0,1)\)
- \((2,-1)\)
- \((6,3)\)
- \((2,1)\)
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The Correct Option is B
Solution and Explanation
the orthocentre of the triangle \(PQR\) is : \((2,-1)\)
Hence, the correct option is (B): \((2,-1)\)
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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.
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