Chords of the curve 4x2+y2-x+4y=0 which subtend right angle at the origin pahrough a fixed point whose coordinates are
Answers
Answer:
Step-by-step explanation:
Answer:
(1/5, 4/5)
Step-by-step explanation:
Hi,
Given equation of the curve is 4x² + y² - x + 4y = 0
Given that chords of the curve subtend right angle at Origin
O(0, 0).
Let the general chord be AB
Let the equation of the line AB be y = mx + c,
On homogenizing the curve equation with the line AB, we get
the equation of pair of straight lines OA and OB
Since y = mx + c
(y - mx)/c = 1
Homogenizing we get
4x² + y² -x(y - mx)/c + 4y(y - mx)/c = 0
4x² + y² - xy/c + mx²/c + 4y²/c - mxy/c = 0
But we know for any pair of straight line to be perpendicular,
Coefficient of x² + Coefficient of y² = 0
⇒ ( 4 + m/c) + ( 1 + 4/c) = 0
⇒ 5 + (m + 4)/c = 0
5c + m + 4 = 0
c + m/5 + 4/5 = 0
On comparing with y = mx + c,
we can observe that y + mx = m/5 + 4/5
m(x - 1/5) + (y - 4/5) = 0
In the above equation , if x= 1/5 and y = 4/5, the equation will
satisfy for all values of m.
Hence, the chords of the curve passes through fixed point
(1/5, 4/5).
Hope, it helps !