Click to Chat

1800-2000-838

+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
```        THREE NORMALS ARE DRAWN FROM THE POINTS (14,7) TO THE CURVE y2-16X-8Y=0. FIND THE COORDINATES OF FEET OF NORMALS.

NOTE. IN EQUNTION 2 IS POWER OF y```
6 years ago

Aman Bansal
592 Points
```										dear manoj,
let the coordinates of foot of perprndicular be (x1,y1)
differentiating the equation to find the slope of the normal,
=> dy/dx = 8/(y-4)
slope = -(dx/dy)
hence we can find the equation of normal in terms of x1 and y1.
then passing through the (x1,y1)
thus we find a cubic equation on solving which we get,
y1= 0, 16 , -4
thus corresponding x1 = 0, 8 , 3
Don’t forget to approve the answer if it is beneficial to you… !!
Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
AMAN BANSAL

```
6 years ago
24 Points
```											Y=mx-2am-am3As we know that this equation implies for standard  parabola But the given equation  is in shifted form After simplifying  the equation (Y-4 )2=16 (X+1)Replacing y-4 =YAnd x+1=1Placing in equation of normal and x=14 and y=7 we get m in cubic find the roots  Remember point of contact also changes.
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 5,900 off
USE CODE: HRITHIK
Get extra Rs. 1,484 off
USE CODE: HRITHIK