Consider f(x)=a/(x-1)^2-1. For which a>1 is the triangle formed by (0,0) and the intersections of f(x) with the positive x- and y-axis isosceles?

We'll first compute these intersections by setting x=0 and y=0 consecutively. This gives y=a-1 and a/(x-1)^2-1=0. Hence we find (x-1)^2=a, so x=1+-sqrt(a). As we have a>1 and we want the intersection with the positive x-axis, we need x=1+sqrt(a). So our two intersections are (0,a-1) and (1+sqrt(a),0).

Now, as the angle at (0,0) is 90 degrees, we need the sides meeting at (0,0) to have the same length, as the third side will always be longer than the other two by the Pythagorean theorem. So, we need d((0,a-1),(0,0))=d((1+sqrt(a)),(0,0)). This gives a-1=1+sqrt(a). Now, as a>1, we can say a=b^2 for some b>0. This gives us b^2-1=1+b and hence b^2-b-2=0. This factorises to (b+1)(b-2)=0. Then, as b>0, we need b=2. This gives us a=4, the required a.

WV
Answered by Ward V. Maths tutor

3287 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you find the integral of sin^2(x) dx?


Differentiate y=sin(x)/5x^3 with respect to x


The line L has equation y = 5 - 2x. (a) Show that the point P (3, -1) lies on L. (b) Find an equation of the line perpendicular to L that passes through P.


Differentiate f(x) = x sin(x)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning