let `A(x_1,0)` and `B(x_2,0)` be the foci of the hyperbola `x^2/9-y^2/16=1` suppose parabola having vertex at origin and focus at `B` intersect the hyperbola at `P` in first quadrant and at point Q in fourth quadrant.

Options

A. `(-(9)/(10),0)`
B. `((2)/(3),0)`
C. `((9)/(10),0)`
D. `((2)/(3),sqrt(6))`

Answer

A

Explanation

Here , `(x^(2))/(9)+(y^(2))/(8)=1`
has foci `(+_ ae, 0)`
where , `a^(2) e^(2)= a^(2)-b^(2)`
`implies a^(2)e^(2)=9-8`
`implies ae=+-1`
`i.e., F_(1),F_(2)=(+-1,0)`

Equation of parabola having vertex O(0,0) and `F_(2)(1,0) ( as ,x_(2) gt 0) `
`y_(2)=4x`
on solving `(x^(2))/(9)+(y^(2))/(8)=1 and y^(2)= 4x,` we get
`x=3//2 and y=+- sqrt(6)`
equation of altitude through M oin `NF_(1) ` is
`(y-sqrt(6))/(x-3//2)=(5)/(2sqrt(6))`
`implies (y-sqrt(6))=(5)/(2sqrt(6))(x-3//2)`. . .(iii)
and equilibrium of altitude through `F_(1) ` is Y=0 . . .(iv)
on solving Eqs. (iii) and (iv) , we get `(-(9)/(10),0)`as orthocentre.