Solveeit Logo
Login

Question

Question: From any point ‘R’, two normal which are right angled to one another are drawn to the hyperbola \[\d...

From any point ‘R’, two normal which are right angled to one another are drawn to the hyperbola x2a2−y2b2=1,(a>b)\dfrac{{{x^2}}}{{a{}^2}} - \dfrac{{{y^2}}}{{{b^2}}} = 1,\left( {a > b} \right). If the feet of the normals are P and Q, then the locus of the circumcentre of the triangle PQR is______. Fill in the blank.
A. x2+y2a2−b2=(x2a2+y2b2)2\dfrac{{{x^2} + {y^2}}}{{{a^2} - {b^2}}} = {\left( {\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}}} \right)^2}
B. x2−y2a2−b2=(x2a2−y2b2)2\dfrac{{{x^2} - {y^2}}}{{{a^2} - {b^2}}} = {\left( {\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}}} \right)^2}
C. x2+y2a2−b2=(x2a2−y2b2)2\dfrac{{{x^2} + {y^2}}}{{{a^2} - {b^2}}} = {\left( {\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}}} \right)^2}
D. x2+y2a2+b2=(x2a2−y2b2)2\dfrac{{{x^2} + {y^2}}}{{{a^2} + {b^2}}} = {\left( {\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}}} \right)^2}

Explanation

Solution

In this question, as it is mentioned that from a single point ‘R’, two normal emerge which are right angles to each other, and form two points P and Q , it is very obvious that tangents at P and Q will intersect at right angles and can be called a cyclic quadrilateral.

Complete step by step answer:
As explained above that tangent at P and Q will intersect at right angles, let that point of intersection be S therefore it will behave as a quadrilateral, PSQR is cyclic. Also, S lies in the direction of the circle of hyperbola, The locus of the point of intersection of the tangents to the hyperbola which meet at right angle, is a circle. This circle is called the "director circle".
S=a2−b2cos⁡θ,a2−b2sin⁡θS = \sqrt {{a^2} - {b^2}} \cos \theta ,\sqrt {{a^2} - {b^2}} \sin \theta
Chord with middle point (h, k) that is circumcentre will be same as equation of chord of contact:
⇒⊥s\Rightarrow \bot s
⇒xha2−ykb2=h2a2−k2b2\Rightarrow \dfrac{{xh}}{{{a^2}}} - \dfrac{{yk}}{{{b^2}}} = \dfrac{{{h^2}}}{{{a^2}}} - \dfrac{{{k^2}}}{{{b^2}}}
Also,
xa2−b2cos⁡θa2−ya2−b2cos⁡θb2=1\dfrac{{x\sqrt {{a^2} - {b^2}\cos \theta } }}{{{a^2}}} - \dfrac{{y\sqrt {{a^2} - {b^2}\cos \theta } }}{{{b^2}}} = 1 are identical, therefore by comparing and solving we get the locus x2+y2a2−b2=(x2a2−y2b2)2\dfrac{{{x^2} + {y^2}}}{{{a^2} - {b^2}}} = {\left( {\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}}} \right)^2}

Therefore, option C is the right answer.

Note: Equation of the director circle of the ellipse is x2+y2=a2−b2{x^2} + {y^2} = a{}^2 - {b^2}, circumcenter of a triangle is defined as it is the point at which the perpendicular bisectors of the sides of a triangle intersect and which is equidistant from the three vertices.