Given: Equation of curve is
4x2+9y2=1 and the tangents to the given curve are perpendicular to the line 2y + x = 0.
Let
(x1,y1) be the point of contact.
The given equation of line 2y + x = 0 can be written as: y = (-1/2) x.
Now by comparing the equation y = (-1/2) x with y = mx + c we get: m = - 1/2 and c = 0
So, the slope of the given line 2y + x = 0 is
m1=−1∕2 As we know that slope of the tangent at any point say
(x1,y1) to a curve is given by:
m=[](x1,y1) Now by differentiating the equation of curve
4x2+9y2=1 with respect to
x we get
⇒8x+18y.=0 ⇒=− ⇒[](x1,y1)=− So, the slope of the tangent is
m2=−4x1∕9y1 It is given that the tangents to the given curve are perpendicular to the line 2y + x = 0.
As we know that, if
L1 and
L2 are two lines with slope
m1 and
m2 respectively and if they are perpendicular to each other then
m1×m2=−1 ⇒m1×m2=−.−=−1 ⇒y1=−x1 ∵ The point
(x1,y1) lies on the curve
4x2+9y2=1 ⇒4x12+9y12=1.
By substituting
y1=−x1 in the above equation we get
⇒4x12+9××x12=1 ⇒x12= ⇒x1=±3∕2√10 When
x1=3∕2√10 then
y1=−x1=−= Similarly when
x1=−3∕2√10 then
y1=−,x1=− = Hence, the points of contact are
(,) and
(,)