 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
Show that all chords of a parabola which subtend a right angle at the vertex pass through a fixed point on the axis of the curve.

```
6 years ago

```							not possible
```
6 years ago
```							Let’s say that the parabola is – y2 = 4ax. Let the ends of the chords on the parabola be A(at12, 2at1) and B(at22, 2at2). Now as they subtend a right angle at the vertex of the parabola - m1m2 = -1, where m1 is slope of line joining AO and m2 is slope of line joining BO, where O = (0,0) or origin. Then, as follows – (at12 – 0) * (at22 – 0) + (2at1 – 0) * (2at2 – 0) = 0, as x1x2 + y1y2 = -1. When we solve this equation we will get the relation between t1 and t2 as follows - t2 = -4 / t1 and let t1 = t, so then the ends of the chords become A(at2, 2at) and B(16a / t2, -8a / t). We can get the equation of the line passing through the ends of chords or just the equation of the chord ‘AB’. On solving the equation of these two points we will arrive at, (hint: factorise denominator of slope) (t – 4/t)y – 2at2 – 8a = 2x – 2at2(2x – 8a) – (t – 4/t)y = 0. This line clearly goes through the point (-4a, 0) which lies on the axis of the parabola y = 0.
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions