A hyperbola having the transverse axis of length $$\sqrt{2}$$ has the same foci as that of the ellipse, $$3x^2 + 4y^2 = 12$$ then this hyperbola does not pass through which of the following points?

We are told that the ellipse is $$3x^2+4y^2=12$$.

First we convert this ellipse to its standard form. Dividing every term by $$12$$ we obtain

$$\frac{x^2}{4}+\frac{y^2}{3}=1.$$

In the standard ellipse $$\dfrac{x^2}{a_e^2}+\dfrac{y^2}{b_e^2}=1$$ we have

$$a_e^2=4,\qquad b_e^2=3.$$

The focal distance for an ellipse is given by the relation

$$c_e^2=a_e^2-b_e^2.$$

Substituting the obtained values,

$$c_e^2=4-3=1\;\;\Longrightarrow\;\;c_e=1.$$

Hence the foci of the ellipse are $$\bigl(\pm1,0\bigr).$$

Because the required hyperbola has the same foci, it must also be centred at the origin with its transverse axis along the $$x$$-axis. Let the hyperbola be

$$\frac{x^2}{a_h^2}-\frac{y^2}{b_h^2}=1.$$

We are further told that the transverse axis of the hyperbola has length $$\sqrt2$$. The transverse axis length equals $$2a_h$$, so we write

$$2a_h=\sqrt2\;\;\Longrightarrow\;\;a_h=\frac{\sqrt2}{2}=\frac1{\sqrt2}.$$

For a hyperbola the focal distance satisfies the well-known formula

$$c_h^2=a_h^2+b_h^2.$$

Here the common foci lie at a distance $$c_h=1$$ from the origin. Substituting the known values,

$$1^2=\left(\frac1{\sqrt2}\right)^2+b_h^2 \;\;\Longrightarrow\;\; 1=\frac12+b_h^2 \;\;\Longrightarrow\;\; b_h^2=1-\frac12=\frac12.$$

Thus the explicit equation of the hyperbola becomes

$$\frac{x^2}{\dfrac12}-\frac{y^2}{\dfrac12}=1.$$

Multiplying numerator and denominator in each fraction by $$2$$ gives

$$2x^2-2y^2=1,$$

or, equivalently,

$$x^2-y^2=\frac12.$$

Now we simply check which of the four given points fails to satisfy this equation.

Option A: $$\left(\dfrac1{\sqrt2},0\right)$$

$$2\!\left(\dfrac1{\sqrt2}\right)^2-2(0)^2 =2\!\left(\frac12\right)-0 =1.$$

The left side equals $$1$$, so the point lies on the hyperbola.

Option B: $$\left(-\sqrt{\dfrac32},1\right)$$

$$2\!\left(\sqrt{\dfrac32}\right)^2-2(1)^2 =2\!\left(\dfrac32\right)-2 =3-2 =1.$$

The equation is satisfied; the point lies on the hyperbola.

Option C: $$\left(1,-\dfrac1{\sqrt2}\right)$$

$$2(1)^2-2\!\left(-\dfrac1{\sqrt2}\right)^2 =2-2\!\left(\dfrac12\right) =2-1 =1.$$

This point also lies on the hyperbola.

Option D: $$\left(\sqrt{\dfrac32},\dfrac1{\sqrt2}\right)$$

$$2\!\left(\sqrt{\dfrac32}\right)^2-2\!\left(\dfrac1{\sqrt2}\right)^2 =2\!\left(\dfrac32\right)-2\!\left(\dfrac12\right) =3-1 =2\neq1.$$

Here the left side is $$2$$, not $$1$$, therefore this point does not lie on the hyperbola.

Hence, the correct answer is Option 4.