Consider the ellipse x^2/9 + y^2/4 = 1 with foci S and S'. Let P be…

Question

Consider the ellipse x^2/9 + y^2/4 = 1 with foci S and S'. Let P be the point of the ellipse lying in the first quadrant at which the two focal radii PS and PS' are perpendicular to each other. The normal to the ellipse at P meets the x-axis at G and the y-axis at H. If the area of triangle OGH (O the origin) equals m/n in lowest terms, find m+n.

Accepted Answer

Here a^2=9, b^2=4, so c^2 = a^2-b^2 = 5, foci at (+-sqrt5, 0).

The locus of points where the two focal radii subtend a right angle is the circle x^2+y^2 = c^2 = 5 (since for a right angle at P with r1+r2=2a and r1^2+r2^2 = (2c)^2, P lies at distance c from the centre). Intersect with the ellipse:
x^2/9 + y^2/4 = 1 and x^2 + y^2 = 5. Solve: from the second y^2 = 5 - x^2. Substitute: 4x^2 + 9(5-x^2) = 36 => -5x^2 +45 = 36 => x^2 = 9/5, y^2 = 16/5. First-quadrant point P = (3/sqrt5, 4/sqrt5).

Normal to the ellipse at (x0,y0): a^2 x / x0 - b^2 y / y0 = a^2 - b^2, i.e. 9x/(3/sqrt5) - 4y/(4/sqrt5) = 5, => 3 sqrt5 x - sqrt5 y = 5.
x-intercept G (y=0): x = 5/(3 sqrt5) = sqrt5/3. y-intercept H (x=0): -sqrt5 y = 5 => y = -sqrt5.

Area of OGH = (1/2)|Gx||Hy| = (1/2)(sqrt5/3)(sqrt5) = (1/2)(5/3) = 5/6.

m/n = 5/6, m+n = 11.
Final answer: 11

Final answer: 11

Consider the ellipse x^2/9 + y^2/4 = 1 with foci S and S'. Let P be…

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