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Options

A. `(a^(2)+b^(2))/(a)`
B. `-((a^(2)+b^(2))/(a))`
C. `(a^(2)+b^(2))/(b)`
D. `-((a^(2)+b^(2))/(b))`

Explanation

Firstly, we obtain the slope fo normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 " at " (a sec theta , b tan theta)`. On differentiating w.r.t.x, we get
`(2x)/(a^(2))-(2y)/(b^(2))xx(dy)/(dx)=0 rArr(dy)/(dx)=(b^(2))/(a^(2))(x)/(y)`
Slope for normal at the point `(a sec theta , b tan theta )` will be
`-(a^(2)b tan theta )/(b^(2)a sec theta)= -(a)/(b) sin theta `
` therefore ` Equation of normal at `(a sec theta, b tan theta )` is
`y-b tan theta= -(a)/(b) sin theta (x-a sec theta)`
`rArr (a sin theta )x + "by"=(a^(2)+b^(2))tan theta`
`rArr ax+b "cosec" theta =(a^(2)+b^(2)) sec theta " ...(i)" `
Similarly, equation of normal to `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` at
`(a sec phi, b tan phi) " is " ax+b y " cosec "phi=(a^(2)+b^(2))sec phi " ... (ii)" `
On subtracting Eq. (ii) from Eq. (i), we get
`b("cosec" theta -"cosec" phi)y=(a^(2)+b^(2))(sec theta-sec phi)`
`rArr y=(a^(2)+b^(2))/(b)*(sec theta-sec phi)/("cosec"theta-"cosec"phi)`
But `(sec theta-sec phi)/("cosec"theta-"cosec"phi)=(sec theta-sec(pi//2-theta))/("cosec"theta-"cosec"(pi//2-theta)) " " [ because phi+theta=pi//2]`
`=(sec theta -"cosec"theta)/(sec theta-sec theta)= -1`
Thus, `y= -((a^(2)+b^(2))/(b)),i.e.k=-((a^(2)+b^(2))/(b))`