2. The tangent to the curve y = ex2 drawn at the point (e, ee) intersects the line joining the points (e - 1, ee-1) and (e+1, ee+1)
3. A hyperbola, having the transverse axis of length 4 sin, is confocal with the ellipse 3x2 + 4y2 =12. Then its equation is ?
To solve question 2, we need to find the equation of the tangent to the curve y = ex^2 at the point (e, ee).
First, let's find the derivative of the function y = ex^2.
Taking the derivative with respect to x, we get:
dy/dx = 2ex
Next, let's substitute x = e into the derivative to find the slope of the tangent at the point (e, ee):
m = dy/dx = 2e * e = 2e^2
Now, we have the slope of the tangent.
Let's find the equation of the tangent using the point-slope form:
(y - ee) = m(x - e)
Substituting the slope (m = 2e^2) and the coordinates of the point (x = e, y = ee), we get:
(y - ee) = 2e^2(x - e)
Simplifying this equation will give us the equation of the tangent.
For question 3, to find the equation of the hyperbola, we need to consider that the hyperbola and ellipse are confocal. This means they share the same foci.
Given that the transverse axis of the hyperbola has a length of 4sin, and since the hyperbola and ellipse are confocal, the length of the transverse axis of the ellipse is also 4sin.
Now, the standard equation of an ellipse with its center at the origin is:
(x^2) / (a^2) + (y^2) / (b^2) = 1
Given that the length of the transverse axis of the ellipse (2a) is equal to 4sin, we have:
2a = 4sin
a = 2sin
We also know that the length of the conjugate axis of the ellipse (2b) is equal to 2sqrt(3) (from the given equation 3x^2 + 4y^2 = 12):
2b = 2sqrt(3)
b = sqrt(3)
Now we have the values of a and b, which we can use to find the equation of the hyperbola.
The standard equation of a hyperbola with its center at the origin is:
(x^2) / (a^2) - (y^2) / (b^2) = 1
Substituting the values of a and b, we get:
(x^2) / (4sin^2) - (y^2) / 3 = 1
This is the equation of the hyperbola.