Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the point where ordinate is 1 such that point lies in the first quadrant?a)√2x - 4y - 4 = 0b)√2x + 4y - 4 = 0c)√2x + 4y + 4 = 0d)None of theseCorrect answer is option 'D'. Can you explain this answer?

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Find the equation of the tangent to the ellipse x²+ 2y² = 4 at the point where ordinate is 1 such that point lies in the first quadrant?

a)
√2x - 4y - 4 = 0
b)
√2x + 4y - 4 = 0
c)
√2x + 4y + 4 = 0
d)
None of these

Correct answer is option 'D'. Can you explain this answer?

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Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the p...

Given

Equation of ellipse is x² + 2y² = 4

Here, we have to find the equation of the tangent to the given ellipse at the point where ordinate is 1 such that the point lies in the first quadrant.

Let the point be P = (c, 1)

Now this point P lies on the ellipse. So, x = c and y = 1 will satisfy the equation x² + 2y² = 4

c/2 + 2 = 4

c = ± √2

So, the point P is (√2, 1) because the point P lies in the first quadrant.

The given equation of ellipse we can be re-written as: (x²/4) + (y²/2) = 1

As we know that, the equation of the tangent to the horizontal ellipse x²/a²+ y²/b² = 1 where 0 < b="" />< a="" at="" the="" point="" />₁, y₁) is given by:

So, the required equation of tangent is √2x + 2y - 4 = 0

Hence, the correct option is (D).

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Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the p...

To find the equation of the tangent to the ellipse x^2 + 2y^2 = 4 at the point where the ordinate is 1 in the first quadrant, we can follow these steps:

Step 1: Find the coordinates of the point on the ellipse where the ordinate is 1:
Given that the ordinate is 1, we can substitute y = 1 into the equation of the ellipse to find the corresponding x-coordinate.
x^2 + 2(1)^2 = 4
x^2 + 2 = 4
x^2 = 2
x = ±√2

Since we are looking for a point in the first quadrant, we take the positive value: (x, y) = (√2, 1)

Step 2: Find the slope of the tangent at this point:
To find the slope of the tangent at a given point on the ellipse, we can differentiate the equation of the ellipse implicitly with respect to x and solve for dy/dx.
Differentiating x^2 + 2y^2 = 4 with respect to x:
2x + 4y(dy/dx) = 0
dy/dx = -2x/(4y) = -x/(2y)

Substituting the coordinates of the point (√2, 1) into the equation, we get:
dy/dx = -√2/(2*1) = -√2/2

Step 3: Use the point-slope form to find the equation of the tangent:
Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
where (x1, y1) is the given point on the tangent and m is the slope of the tangent.

Substituting the values into the equation, we get:
y - 1 = (-√2/2)(x - √2)

Simplifying the equation:
y - 1 = -√2/2 * x + √2/2 * √2
y - 1 = -√2/2 * x + √2
y = -√2/2 * x + √2 + 1
y = -√2/2 * x + √2 + 2/2
y = -√2/2 * x + √2 + 1
y = -√2/2 * x + √2 + 1

Comparing the equation with the options given, we can see that none of the options match the equation obtained. Therefore, the correct answer is option 'D' (None of these).

Answered by Wizius Careers · View profile

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Find the equation of the tangent to the ellipse x2 + 2y2 = 4 at the point where ordinate is 1 such that point lies in the first quadrant?a)√2x - 4y - 4 = 0b)√2x + 4y - 4 = 0c)√2x + 4y + 4 = 0d)None of theseCorrect answer is option 'D'. Can you explain this answer?

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