If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse (x^2)/40+(y^2)/10=1 is x^2+y^22=ax^2+by^2, then a–b is…

Question

If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse (x^2)/40+(y^2)/10=1 is x^2+y^22=ax^2+by^2, then a–b is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) 30. Slope of OP =k/h⇒ slope of the tangent is (-h)/k Hence, the equation of tangent PQ is ((y-k))/(x-h)=(-h)/k ⇒ y=-h/kx+(h^2+k^2)/k ⇒h2+k2k^2=40-hk^2+10 condition of tangency is c^2=a^2m^2+b^2 ⇒x^2+y^22=40x^2+10y^2 ⇒a–b=30

If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is ${(x^{2} + y^{2})}^{2} = a x^{2} + b y^{2}$ , then $(a - b)$ is equal to

Solution

More Ellipse Questions

If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse x240+y210=1 is x2+y22=ax2+by2, then a–b is equal to

Solution

More Ellipse Questions

If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse $\frac{x^{2}}{40} + \frac{y^{2}}{10} = 1$ is ${(x^{2} + y^{2})}^{2} = a x^{2} + b y^{2}$ , then $(a - b)$ is equal to